LIBRARY 


UNIVERSITY  OF  NORTH  CAROLINA 


Alcove  Shelf 


UNIVERSITY  OF  N.C.  AT  CHAPEL  HILL 


00011546124 


CHEMICAL         AnJ^^y 
PEOBLEMS  AND  REACTIONS, 

TO   ACCOMPANT 

STOCKHARDT'S  ELEMENTS  OF  CPIEMISTRY. 


BY 

JOSIAH    P.'^OOKE,  Jr. 


#H:?-L:i3fary 

il;^  Vnhimliy  of  North  Ca  o;;;.. 


CAMBEIDGE: 

PUBLISHED    BY    JOHN    BAETLETT, 

ISoofeseller  to  tje  Sinfbcrsitj). 

1857. 


C  A  31  B  E  I  D  G  E  : 

ELECTROTTPED  AND  PRINTED  BY  METCALF  AND   COMPANY. 


PREFACE. 


This  book  has  been  prepared  solely  for  the  use  of  the 
undergraduates  of  Harvard  College.  It  contains  a  collection 
of  chemical  problems  and  reactions,  with  references  to  the  sec- 
tions of  Stockhardt's  Elements  of  Chemistry,  and  also  a  few 
chapters  on  the  chemical  nomenclature  and  the  use  of  chemical 
symbols,  subjects  which  are  not  sufficiently  developed  in  that 
text-book  for  the  purposes  of  college  instruction.  In  writing 
chemical  symbols  the  author  has  adopted  a  uniform  system 
throughout  the  volume,  which,  as  he  hopes,  will  be  found  to 
be  at  once  expressive  and  clear.  The  problems  and  reactions 
cover  the  Inorganic  portion  of  Stockhardt's  Elements  ;  the 
problems  have  only  been  extended  to  the  section  on  the  Heavy 
Metals.  Beyond  this,  the  reactions  alone  have  been  given,  as  it 
was  supposed  that,  before  reaching  this  section,  the  student 
will  easily  be  able  to  propose  problems  for  himself.  In 
solving  many  of  the  problems  it  will  be  found  convenient  to 
use  logarithmic  tables  of  four  places,  which,  with  several  other 
tables,  will  be  found  at  the  end  of  the  volume.  The  student 
is  advised  to  remove  the  tables  of  logarithms,  and  paste  them 
for  use  on  a  card.  The  difficulty  of  insuring  complete  accu- 
racy in  the  printing  of  chemical  formulge  can  be  known  only 
to  those  who  have  had  to  see  a  book  of  this  kind  through  the 
press.  Several  errors  have  been  already  discovered,  and 
corrected,  but  others  unquestionably  exist. 

Cambeidge,  May  I5th,  1857. 


Digitized  by  the  Internet  Archive 

in  2012  with  funding  from 

University  of  North  Carolina  at  Chapel  Hill 


http://archive.org/details/chemicalproblemsOOcook 


NOMENCLATURE  OF   CHEMISTRY. 


Origin  of  the  Nomenclature.  —  Previous  to  the  year  1787 
the  names  given  by  chemists  or  alchemists  to  substances 
were  not  conformed  to  any  general  rules.  Many  of  these  old 
names,  such  as  Oil  of  Vitriol,  Calomel,  Corrosive  Sublimate, 
Red-Precipitate,  Saltpetre,  Liver  of  Sulphur,  Cream  of  Tartar, 
Glauber^s  and  Epsom  Salts,  are  stUl  retained  in  common  use. 
As  chemical  science  advanced,  and  the  number  of  known 
substances  iucreased,  it  became  important  to  adopt  a  scientific 
nomenclature.  The  admirable  system  now  in  use  is  due  al- 
most entirely  to  Lavoisier,  who  reported  to  the  French  Acad- 
emy on  the  subject,  in  the  name  of  a  committee,  in  1787. 
This  system,  now  known  as  the  Lavoisierian  nomenclature, 
was  generally  adopted  by  scientific  men  soon  after  its  pub- 
lication, and  has  not  been  materially  modified  since.  In  it 
the  name  of  a  substance  is  made  to  indicate  the  composi- 
tion. 

Names  of  the  Elements.  —  The  names  of  the  elements  are 
the  only  ones  which  are  now  independent  of  any  rule.     Those 
which  were  known  before  the  adoption  of  the  nomenclature, 
1 


2  NOMENCLATURE    OF   CHEMISTRY. 

such,  as  Sulphur,  Phosphorus,  Iron,  Lead,  retain  their  old 
names.  Several  of  the  more  recently  discovered  elements 
have  been  named  in  allusion  to  some  prominent  property  or 
some  circumstance  connected  with  their  history;  as,  Oxygen 
from  o^vs,  yei/mc*  (acid-generator)  ;  Hydrogen,  from  vbap,  yewaca 
(water-generator) ;  Chlorine,  from  -xkapos  (green) ;  Iodine, 
from  labrjs  (violet) ;  Bromine,  from  ^papos  (fetid  odor),  &c. 
The  names  of  the  newly  discovered  metals  have  a  common 
termination,  um,  as  Platinum,  Potassium,  Sodium ;  and  the 
names  of  a  class  of  the  metalloids  terminate  in  ine,  as  Chlo- 
rine, Bromine,  &c. ;  but  except  in  these  respects  the  names 
of  the  elements  are  entirely  arbitrary. 

Classification  of  Compounds.  —  There  are  three  orders  of 
chemical  compounds :  —  1st,  Binary  Compounds,  consisting  of 
two  elements,  or  of  the  representatives  of  two  elements ;  2d, 
Ternary  Compounds,  consisting  of  three  elements,  or  of  their 
representative ;  and  Sd,  Quaternary  Compounds,  consisting  of 
four  elements,  or  their  representatives.  There  are  some  chem- 
ical compounds  containing  more  than  four  elements ;  but  in 
most  cases  two  or  more  of  these  elements  are  representatives, 
i.  e.  occupy  the  place,  of  only  one,  as  will  be  explained  farther 
on.  Binary  compounds  are  subdivided  into  two  classes,  Elec- 
tro-Positive Binaries,  or  Bases,  and  Electro-Negative  Binaries, 
or  Acids.  Each  of  these  classes  is  distinguished  by  a  peculiar 
set  of  properties,  or  at  least  this  is  the  case  with  the  promi- 
nent members  of  either  class ;  but  the  two  classes  merge  so 
gradually  into  each  other,  that  it  is  impossible  to  draw  a  line 
of  demarcation  between  them ;  and  there  is  a  large  class  of 
intermediate  compounds,  which  either  partake  of  the  proper- 


NOMENCLATURE    OF   CHEMISTRY.  3 

ties  of  both,  or  are  entirely  indifferent.  Indeed,  the  binary 
compounds  may  best  be  regarded  as  forming  a  continuous  series 
of  substances,  varying  in  their  properties  from  those  of  strong 
acids  on  the  one  hand  to  those  of  strong  bases  on  the  otlier, 
and  with  every  possible  grade  of  qualities  between  the  two 
extremes.  In  this  series  each  binary  may  be  considered  as 
an  electro-positive  compound  or  base  towards  all  those  which 
precede  it,  and  as  an  electro-negative  compound  or  acid  to- 
wards all  those  which  follow  it.  Ternary  compounds  are  gen- 
erally, at  least  in  Inorganic  Chemistry,  composed  of  two 
binaries,  i.  e.  of  an  acid  and  a  base,  and  are  then  called 
Salts.  The  quaternary  compounds  are  generally  composed  of 
two  salts,  and  are  called  Double  Salts. 

Names  of  Binaries.  —  The  most  important  binaries,  as  well 
as  those  which  have  been  the  best  studied,  are  the  compounds 
of  oxygen  with  the  other  elements.  To  these  the  generic 
term  Oxide  has  been  applied.  The  electro-positive  binaries 
are  called  simply  Oxides  of  the  elements  of  which  they  con- 
sist.   Thus  we  have 

Oxide  of  Hydrogen,  consisting  of  oxygen  and  hydrogen. 
Oxide  of  Potassium,         "  "  "  potassium. 

Oxide  of  Sodium,  "  "  sodium. 

When  the  name  of  the  metal  ends  in  iim,  the  name  of  the 
compound  with  oxygen  is  frequently  formed  by  changing  this 
termination  into  a,  with  such  other  modifications  of  the  termi- 
nal letters  as  euphony  may  require.     Thus  we  use,  instead  of 

Oxide  of  Sodium,  Soda. 

Oxide  of  Potassium,  Potassa. 


4  NOMENCLATURE    OF    CHEMISTRT. 

Oxide  of  Calcium,  Calcia  (or  Lime).* 

Oxide  of  Barium,  Baryta. 

Oxide  of  Strontium,  Strontia. 

Oxide  of  Magnesium,  Magnesia. 

Oxide  of  Aluminum,  Alumina. 

The  two  names  are  in  all  cases  synonymous.  Generally 
oxygen  combines  with  an  element  in  more  than  one  propor- 
tion ;  then,  in  order  to  distinguish  between  the  different  oxides 
of  the  same  element,  we  use  various  Latin  and  Greek  prefixes, 
such  as  suh,  proto,  sesqui,  deuto,  hyper.  This  is  well  illustrated 
by  the  names  of  the  different  oxides  of  mercury  and  man- 
ganese, which  are  as  follows. 

Composition. 

Names. 

Suho:side  of  Mercury 
Protoxide  of  Mercury 

Protoxide  of  Manganese 
Sesquioxide  of  Manganese 
£li/pero:side  of  Manganese 

The  electro-negative  binary  compounds  of  oxygen  (the  acids) 
are  named  on  a  different  principle.  These  are  called  different 
kmds  of  acids.  If  the  element  forms  but  one  acid  with  oxy- 
gen, the  name  is  formed  by  adding  to  the  name  of  the  acid  the 
termination  ic,  with  such  changes  of  the  final  letters  as  eu- 
phony may  require.     Thus,  carbon  and  oxygen  form  Carbonic 

*  The  common  name  Lime  is  much  more  frequently  used  than  either  of  its 
scientific  synonymes,  Oxide  of  Calcium,  or  Calcia.  Indeed,  the  last  has  never 
been  in  general  use. 


Mercury. 

Oxygen. 

100 

4 

100 

8 

Manganese. 

Oxygen. 

27.6 

8 

27.6 

12 

27.6 

16 

Arsenic. 

Oxygen. 

75 

24 

75 

40 

Phosphorus. 

Oxygen. 

32 

24 

32 

40 

NOMENCLATURE    OF    CHEMISTRY.  0 

Acid.  When  the  element  forms  two  acids,  by  combining  with 
different  amounts  of  oxygen,  the  termination  ie  is  reserved  for 
that  containing  the  most  oxygen,  while  the  termination  ous  is 
given  to  the  other.     We  have,  for  example, 

ArsenioMs  Acid 
Arsem'c  Acid 

Phosphorotw  Acid 
Phosphorzc  Acid 

.  If  oxygen  combines  with  an  element  in  more  than  two  pro- 
portions, to  form  acids,  the  names  are  formed  with  the  Greek 
prefix  Jiypo,  indicating  a  less,  or  the  Latin  prefix  per,  indicating 
a  greater,  amount  of  oxygen  than  that  contained  in  the  acids 
to  whose  names  they  are  prefixed.  The  acid  compounds  of 
sulphur  and  oxygen  are 

Sulphur.  Oxygen. 

IIyposVi\]Amvous  Acid  16  8 

Sulphuroits  Acid  16  16 

Hypo&\}\^\mxic  Acid  16  20 

Sulphur^c  Acid  16  24 

The  acid  compounds  of  chlorine  and  oxygen  are 

Chlorine.  Oxj'gen. 

IIypoQh\.OYous  Acid  35.5  8 

Chlovous  Acid 
Jlypochloric  Acid 
Chloric  Acid 
PerdiAoTic  Acid 

Very  frequently  the  higher  degrees  of  oxidation  of  an  ele- 
1* 


35.5 

24 

35.5 

.32 

35.5 

40 

35.5 

56 

Nitrogen. 

14 

Oxygi 
8 

14 

16 

14 

24 

14 

32 

14 

40 

b  NOMENCLATURE   OF   CHEMISTRY. 

ment  are  acids,  when  the  lower  degrees  are  bases,  or  indifferent 
compounds.  This  is  the  case  with  the  oxides  of  manganese. 
Besides  the  three  ah-eady  mentioned,  there  are  also 

Manganese.  Oxygen. 

Manganic  Acid  27.6  24 

Permangamc  Acid  27.6  28 

The  different  oxides  of  nitrogen  are,  in  like  manner. 

Protoxide  of  Nitrogen 

Deutoxide  of  Nitrogen 

Nitro^<s  Acid 

Hi/ponitroits.  Acid 

Nitr^'c  Acid 

It  will  be  noticed  that  here,  as  m  other  places,  the  term 
oxides  is  used  genericaUy  for  all  the  compounds  of  an  element 
with  oxygen,  whether  acids  or  bases,  and  the  terms  protoxide, 
&c.  to  designate  the  specific  compounds.  It  is  not  unfre- 
quently  the  case  that  the  same  oxide  acts  as  an  acid  under 
some  circumstances,  and  as  a  base  under  others,  and  accord- 
ingly is  known  under  two  different  names.  "Water,  when  a 
base,  is  called  Oxide  of  Hydrogen,  but  when  an  acid,  it  is 
named  Hydric  Acid.  The  oxide  of  aluminum,  when  a  base, 
is  called  Sesquioxide  of  Aluminum,  but  when  an  acid,  Alu- 
minic  Acid. 

Next  to  the  oxides,  the  compounds  of  sulphur  with  the 
elements  are  the  most  important  binaries.  These  are  called 
sulphides,  and,  as  a  general  rule,  for  every  oxide  there  is  a 
corresponding  sulphide,  which  is  named  after  the  analogy  of 
the  name  of  the  oxide.  Thus  we  have  two  sulphides  of  mer- 
cury :  — 


NOMENCLATURE   OF   CHEMISTRY. 


/SMfeulphide  of  Mercury 
Protosulphide  of  Mercury. 

There  ai'e  three  sulpliides  of  iron :  - 

Protosulphide  of  Iron 
Sesquisulphide  of  Iron 
Persulphide  of  Iron 

The  last  corresponds  to  no  oxide.  The  sulphides,  like  the 
oxides,  may  be  divided  into  basic  or  indifferent  sulphides,  and 
into  acid  sulphides.  The  sulphur  bases  are  named  as  above. 
The  sulphur  acids  have  been  named  by  prefixing  to  the  name 
of  the  oxygen  acid  the  letters  sulpho.  Thus  we  have,  cor- 
responding to 


[ercury. 
100 

Sulphur. 
8 

100 

16 

Iron. 
28 

Oxygen. 
8 

26 

12 

28 

16 

Arsenious  Acid, 
Arsenic  Acid, 
Carbonic  Acid, 
Hydric  Acid, 


SiilphoBXsemons  Acid. 
Sulphoarsenic  Acid. 
iS^M^^ocarbonic  Acid. 
^M^Aohydric  Acid. 


The  names  of  the  binary  compounds  of  the  other  elements 
are  named  after  the  same  analogy  as  the  oxides  and  sulphides. 
Thus  with  the  other  elements 


Oxygen 

Fluorine 

Chlorine 

Bromine 

Iodine 

Sulphur 

Selenium 


forms 


Oxides. 

Fluorides. 

Chlorides. 

Bromides. 

Iodides. 

Sulj)hides. 

Selenides. 


Q  NOMENCLATURE    OF    CHEMISTRY. 

Tellurium  forms  Tellurides. 

Nitrogen  "  Nitrides. 

Phosphorus  "  Phosphides. 

Arsenic  "  Arsenides,  &c. 

"We  distinguish  the  different  fluorides,  bromides,  &c.  by 
prefixes,  as  we  do  the  oxides.  Each  of  these  classes  of  com- 
pounds may  be  subdivided,  like  the  oxides  or  sulphides ;  but 
only  in  a  few  cases  has  the  nomenclature  of  the  oxygen  and 
sulphur  acids  been  extended  to  them.  Almost  the  only 
instances  are  the  names  of  the  compounds  of  the  first  few 
elements  of  the  last  list  with  hydrogen,  which  are  some- 
times called  Fluohydric,  Chlorohydric,  Bromohydric  Acids, 
&c.',  but  even  here  the  synonymous  names  Hydrofluoric, 
Hydrochloric,  Hydrobromic  Acids,  &c.  are  more  generally 
used. 

Names  of  Ternaries.  —  The  name  of  a  ternary  compound, 
or  salt,  is  very  simply  formed  from  the  names  of  the  acid  and 
base  of  which  it  consists.  The  termination  of  the  name  of 
the  acid  is  changed,  if  ic,  into  ate ;  if  ous,  into  ite  ;  and  the 
name  of  the  base  added.  Thus,  sulphur^c  acid  and  oxide  of 
sodium  form  Sulphate  of  the  Oxide  of  Sodium  (or  of  Soda) ; 
sulphurot<s  acid,  Sulph^^e  of  the  Oxide  of  Sodium.  In  like 
manner,  hyposulphurzc  acid  forms  Hyposulphafe,  and  hypo- 
sulphuroMS  acid,  Hyposulph^Ve  of  the  Oxide  of  Sodium.  So, 
also,  Chlorite  of  Oxide  of  Potassium,  and  Chlorate  of  Oxide 
of  Potassium ;  Selenite  of  Oxide  of  Barium,  and  Selenate  of 
Oxide  of  Barium  ;  Nitrite  of  Oxide  of  Calcium,  and  Nitrate  of 
Oxide  of  Calcium ;  Arsenite  of  Oxide  of  Lead,  and  Ai'senate 
of  Oxide  of  Lead.     When  the  base  is  an  oxide  of  one  of  the 


NOMENCLATURE    OF   CHEMISTRY.  9 

heavy  metals,  the  name  of  the  salt  is  frequently  abbreviated 
by  leaving  out  the  word  oxide  ;  as,  Sulphate  of  Lead,  for  Sul- 
phate of  the  Oxide  of  Lead ;  and  Nitrate  of  Copper,  for 
Nitrate  of  the  Oxide  of  Copper.  When  the  base  is  an  oxide 
of  one  of  the  light  metals,  by  substituting  for  Oxide  of  So- 
dium, Oxide  of  Potassium,  &c.,  the  single  words  Soda,  Po- 
tassa,  &c.,  as  already  explained  on  page  3.  Listead  of  Sul- 
phate of  Oxide  of  Barium,  chemists  more  frequently  write 
the  shorter  Sulphate  of  Baryta;  instead  of  Nitrate  of  Oxide 
of  Calcium,  Nitrate  of  Lime.  "When  the  same  acid  combines 
with  two  or  more  different  oxides  of  the  same  element  to  form 
salts,  these  are  distinguished  by  introducing  the  specific  name 
of  the  oxide  iato  the  name  of  the  salt ;  as.  Sulphate  of  Prot- 
oxide of  Iron,  and  Sulphate  of  AS'es§'M^oxide  of  Iron ;  Nitrate 
of  ASwSoxide  of  Mercury,  and  Nitrate  of  Protoxide  of  Mer- 
cury. It  is  frequently  the  case  that  an  acid  combines  with 
the  same  base  in  two  or  more  different  proportions,  forming 
two  or  more  different  salts.  In  order  to  distinguish  these 
salts,  the  one  is  called  the  neutral  salt ;  that  containing  more 
base  than  the  neutral  salt,  a  hasic  ;  and  that  containing  less,  an 
acid.    Thus  we  have 

Basic  Phosphate  of  Lime. 
Neutral  Phosphate  of  Lime. 
Acid  Phosphate  of  Lime. 


So,  also, 


Basic  Chromate  of  Lead. 
Neutral  Chromate  of  Lead. 


When  there  are  several  basic  salts,  they  are  distinguished 
by  Latin  prefixes.    Thus  there  are  five  acetates  of  lead  :  — 


10  NOMENCLATURE    Or   CHEMISTRT. 


Neutral  Acetate  of  Lead 

Bibasic  Acetate  of  Lead 

Sesquibasic  Acetate  of  Lead 

Tribasic  Acetate  of  Lead 

Sexbasic  Acetate  of  Lead 
The  acid  salts  are  distinguished  by  placing  the  Latin  prefixes 
directly  before  the  name  of  the  neutral  salts  ;  as, 

Soda.  Boracic  Acid. 

Borate  of  Soda  31  35 

Biborate  of  Soda  31  70 


Oxide  of  Lead. 

Acetic  Acid. 

112 

51 

224 

51 

280 

51 

336 

51 

672 

51 

Potassa. 

Chromic  Acid. 

Neutral  Chromate  of  Potassa 

47 

51 

Bichromate  of  Potassa 

47 

102 

Trichromate  of  Potassa 

47 

153 

Most  of  the  oxygen  acids  are  commonly  used  only  when  in 
combination  with  water.  These  compounds  are  true  salts,  in 
which  the  water  plays  the  part  of  a  base.  The  common 
Sulphuric  Acid  is  a  Sulphate  of  Water,  and  the  common 
Nitric  Acid,  Nitrate  of  Water.  Properly  speaking,  the  terms 
Sulphuric  and  Nitric  Acid  ought  only  to  be  applied  to  the 
anhydrous  compounds ;  but  custom  authorizes  us  to  extend 
these  names  to  the  hydrates. 

The  names  of  the  ternary  sulphur  compounds,  the  sulphur 
salts,  are  formed  in  the  same  way  as  those  of  the  oxygen  salts ; 
thus,  the  compound  of  sull^hocarbon^c  acid  and  sulphide  of 
sodium  is  called  the  Sulphocarbona^e  of  the  Sulphide  of  So- 
dium ;  the  compound  of  sulphoarsen^c  acid  and  sulphide  of 
potassium,  the  Sulphoarsenafe  of  the  Sulphide  of  Potassium ; 
and  the  compound  of  sulphoarsenioz<s  acid  and  the  same  base, 
SulphoarsenzVe  of  the  Sulphide  of  Potassium. 


NOMENCLATURE    OF    CHEMISTRY.  11 

The  names  of  the  ternary  fluorine,  chlorine,  bromine,  and 
iodine  compounds  are  formed  after  a  different  analogy.  These 
are  generally  regarded  as  double  salts,  and  are  called  the 
double  fluorides,  chlorides,  bromides,  or  iodides  of  the  two 
metals  which  they  contain.     Hence  the  names 

Double  Chloride  of  Aluminum  and  Sodium. 

Double  Chloride  of  Platinum  and  Ammonium. 

Names  of  Quaternaries.  —  The  double  salts  formed  from 
two  salts  each  contaiaing  the  same  acid  but  different  bases, 
are  called  double  salts  of  the  two  bases.  Thus,  the  compound 
of  sulphate  of  alumina  and  sulphate  of  potassa  is  called  the 
Double  Sulphate  of  Alumina  and  Potassa ;  the  compound  of 
sulphate  of  soda  and  sulphate  of  zinc,  the  Double  Sulphate 
of  Soda  and  Zinc.  So,  also,  the  Double  Sulphate  of  Potassa 
and  Magnesia,  and  the  Double  Sulphate  of  Potassa  and  Water. 
The  last  is  more  frequently  called  Bisulphate  of  Potassa,  as  if 
it  were  composed  of  one  equivalent  of  base  and  two  equiva- 
lents of  acid,  instead  of  being,  as  it  probably  is,  a  compound 
of  two  salts.  Similar  incongruities  in  the  nomenclature,  aris- 
ing frequently  from  different  theories  in  regard  to  the  consti- 
tution of  substance,  are  not  uncommon  among  the  higher  orders 
of  compounds. 

The  Lavoisierian  nomenclature  has  not  been  found  suffi- 
ciently expansive  to  meet  the  requirements  of  modern  sci- 
ence, not  only  on  account  of  the  complex  character  of  many 
of  the  newly  discovered  compounds,  but  more  especially 
because  it  involves  the  ideas  of  a  peculiar  theory,  which, 
although  once  almost  universally  received,  is  not  now  so  gen- 
erally admitted.     The  inadequacy  of  the  old  nomenclature  to 


12  NOMENCLATUKE    OF   CHEMISTRY. 

afford  names  for  the  great  variety  of  chemical  compounds,  or 
to  describe  the  infinite  changes  to  which  they  are  hable,  has 
given  rise  to  a  peculiar  chemical  language,  analogous  to  the 
language  of  mathematics,  called  Chemical  Symbols.  To  ex- 
plain the  use  of  this  language  will  be  the  object  of  the  next 
chapter. 


CHEMICAL    SYMBOLS. 


Since  all  matter  is  composed  of  one  or  more  of  sixty-two 
different  substances,  never  as  yet  decomposed,  and  hence  called 
Elements,  it  is  evident  that,  if  we  adopt  an  arbitrary  symbol  for 
each  of  these  elements,  we  shall  be  able,  by  combining  them  to- 
gether, to  express  aU  possible  varieties  of  combination.  More- 
over, since  the  elements  always  combine  in  certain  fixed  and 
definite  proportions  by  weight,  it  is  equally  evident  that,  if  we 
assign  to  each  of  these  symbols  a  certain  weight,  we  shall  be 
able  to  indicate  the  relative  quantities  of  the  difierent  elements 
which  enter  into  any  compound. 

Symbols  of  Elements.  —  It  has  been  agreed  by  chemists  of 
different  nations  to  use,  as  symbols  of  the  elements,  the  first 
letters  of  their  Latin  names.  "When  two  or  more  names  com- 
mence with  the  same  letter,  a  second  letter  is  added  for  dis- 
tinction. The  first  letter  is  printed  or  written  in  capitals,  and 
the  second,  when  used,  in  small  letters,  immediately  following 
the  first.  A  list  of  the  Elements,  with  their  Symbols,  is  given 
on  the  following  page. 
2 


14 


CHEMICAL   SYMBOLS. 


CHE]\nCAL  SYIVEBOLS  A2TD  EQUIVALENTS. 


Aluminum 

Al  = 

13.7 

Xickel 

Ni  = 

29.6 

Antimony  (Stibium) 

Sb  = 

129 

^Niobium 

Nb 

Arsenic 

As  = 

75 

Nitrogen 

N  = 

14 

Barium 

Ba  = 

68.5 

Xorium 

No 

Bismuth 

Bi  = 

213 

Osmium 

03  = 

99.6 

Boron 

B    = 

10.9 

Oxygen 

0    = 

8 

Bromine 

Br  = 

80 

Palladium 

Pd  = 

53.3 

Cadmium 

Cd  = 

56 

Pelopium 

Pe 

Calcium 

Ca  = 

20 

PJios2)horus 

P    = 

32 

Carbon 

C    = 

6 

Platinum 

Pt  = 

98.7 

Cerium 

Ce  = 

47 

Potassium  (Kalium) 

K  = 

39.2 

Chlorine 

CI  = 

35.5 

Rhodium 

E   = 

52.2 

Chromium 

Cr  = 

26.7 

Euthenium 

Eu  = 

52.2 

Cobalt 

Co  = 

29.5 

vSelenium 

Se  = 

89.5 

Cop/>er  (Cuprum) 

Cu  = 

31.7 

Silicon 

Si  = 

21.3 

Didymium 

D 

Silver  (Argentum) 

Ag  = 

lOS.l 

Erbium 

E 

Sodium  (Katrimn) 

Na  = 

23 

Fluorine 

Fl  = 

18.9 

Strontium 

Sr  = 

43.8 

Glucinum 

G   = 

4.7 

Sulphur 

S   = 

16 

Gold  (Aurum) 

Au  = 

197 

Tantalum 

Ta  = 

184 

Eydrogen 

H    = 

1 

Tellurium 

Te  = 

64.2 

Iodine 

I     = 

127.1 

Terbium 

Tb 

Iridium 

Ir   = 

99 

Thorium 

Th  = 

59.6 

Iron  (Ferrum) 

Fe  = 

2S 

Tin  ( Stannum) 

Sn  = 

59 

Lanthanium 

La 

Titanium 

Ti  = 

25 

Lead  (Plumbum) 

Pb  = 

103.7 

Tungsten  ( 'Wolfram) 

W  = 

95 

Lithium 

Li  = 

6.5 

Uranium 

U  = 

60 

Magnesium 

Mg  = 

12.2 

Vanadium 

V  = 

68.6 

Manganese 

Mn  = 

27.6 

Yttrium 

Y 

Mercury  (Hydrargyrum^ 

Hg  = 

100 

Zinc 

Zn  = 

32.6 

Molybdenum 

Mo  = 

46 

Zirconium 

Zr  = 

22.4 

CHEMICAL    SYMBOLS.  15 

The  student  will  do  well  to  notice  in  the  foregoing  list  the 
symbols  of  those  elements  whose  Latin  names  commence  with 
letters  differing  from  the  initial  letters  of  the  English  names, 
since  they  are  not  so  easily  remembered  as  the  others. 

Chemical  Equivalents.  —  The  chemical  symbols  not  only 
stand  for  the  names  of  the  elements,  but  also  for  a  fixed  pro- 
portional weight  of  each.  These  weights  are  given  in  the 
above  table,  opposite  to  the  symbols.  They  have  only  rela- 
tive values ;  if  one  is  in  pounds,  all  the  rest  are  in  pounds ; 
and  if  one  is  in  ounces,  all  the  rest  are  in  ounces.  We  may 
leave  the  standard  indefinite,  and  express  the  weight  in  parts  ; 
then  Al  stands  for  13.7  parts  of  aluminum ;  Sb  stands  for  129 
parts  of  antimony,  &c.  The  weight  of  an  element  indicated 
by  its  symbol  is  called  one  equivalent,  and  it  is  a  law  of 
chemistry  that  elements  always  combine  by  equivalents ;  that 
is,  one  equivalent  of  one  combines  "with  one  equivalent  of  an- 
other, or  else  several  equivalents  of  one  combine  with  one  or 
with  several  equivalents  of  another. 

As  stands  for  75   parts   of  Arsenic,  or  one  equivalent. 
BK         "  68.5         "       Barium,  «  " 

H  "  1  "       Hydrogen,       «  « 

0  "  8  «       Oxygen,  "  « 

In  order  to  express  two,  three,  or  more  equivalents  of  an 
element,  we  place  a  figure  just  below  the  symbol  at  its  right 
hand;  thus, 

02  stands  for  16  parts  of  Oxygen,  or  two  equivalents. 

03  "         24  "  "  three  equivalents. 

These  figures  merely  multiply  the  symbols  beneath  which 


16  CHEMICAL    SYMBOLS. 

they  stand,  and  must  not  be  confounded  with  algebraic  powers, 
which  are  sometimes  written  in  a  similar  way.  We  sometimes 
place  the  figure,  though  larger,  before  the  symbol ;  2  0  means 
exactly  the  same  thing  as  Oj. 

Symbols  of  Compounds.  —  In  order  to  form  the  symbol  of  a 
compound,  Ave  write  the  symbols  of  the  elements  of  which  it 
consists  one  after  the  other,  indicating  by  means  of  figures  the 
number  of  equivalents  of  each  which  have  entered  into  com- 
bination. Thus,  H  0  is  the  symbol  of  water,  a  compound  con- 
sisting of  one  equivalent  or  one  part  of  hydrogen,  and  of  one 
equivalent  or  eight  parts  of  oxygen ;  S  O3  is  the  symbol  of 
sulphuric  acid,  a  compound  consisting  of  one  equivalent  or  six- 
teen parts  of  sulphur,  and  of  tliree  equivalents  or  twenty-four 
parts  of  oxygen ;  C12  Hn  On  is  the  s^bol  of  common  sugar, 
a  compound  consisting  of  twelve  equivalents  of  carbon,  eleven 
equivalents  of  hydrogen,  and  eleven  equivalents  of  oxygen. 

Binary  Compounds.  —  The  symbols  of  binary  compounds 
are  formed  by  writing  the  symbols  of  the  two  elements  to- 
gether, taking  care  to  place  the  symbol  of  the  metal,  or  of 
the  most  electro-positive  element,  first.  The  binary  symbol  thus 
obtained  represents  always  one  equivalent  of  the  compound. 
The  weight  of  this  equivalent  is  evidently  the  sum  of  the 
weights  of  the  equivalents  of  the  elements  entering  into  the 
compound. 

14  +  40  =  54. 

N  O5   stands  for  one  equiv.  or  54  parts  of  Nitric  Acid. 

16  +  24  =  40. 

S  O3  "  "  40        «       Sulphuric  Acid. 


CHEMICAL    SYMBOLS.  17 

6  +  16  =  22. 

C  O2  stands  for  one  equiv.  or  22  parts  of  Carbonic  Acid. 

1    +   8   :=    9. 

II 0  «  «  9        «        Water. 

28  +  8  =  36. 

FeO  "  «  36        "        Oxide  of  Iron. 

20  +  8  =  28. 

CaO  "  "  28        «        Lime. 

23  +  8  =  31. 

Na  0         "  "  31        «        Oxide  of  Sodium. 

6  +  32  =  38. 

CS2  "  "  38        "        Sulphocarbonic  Acid. 

75+  48  =  123. 

As  S3  "  123        "        Sulphoarsenious  Acid. 

28  +  16  =  44. 

FeS  "  "  44        «        Sulphide  of  Iron. 

39  +  16  =  55. 

KS  "  "  55        "        Sulphide  of  Potassium. 

In  order  to  express  two  or  more  equivalents  of  a  binary,  we 
place  a  figure  immediately  before  the  symbol,  like  an  algebraic 
coefficient.  A  figure  so  placed  always  multiplies  the  whole 
hinary. 

3  S  O3  stands  for  3  equiv.  or  120  parts  of  Sulphuric  Acid. 
5PbO        "        5        "        560       «        Oxide  of  Lead. 

Ternary  Compounds.  —  The  symbol  of  a  ternary  compound 
is  formed  by  writing  together  the  symbols  of  the  two  binaries 
of  which  it  consists,  separated  by  a  comma,  taking  care  to 
place  the  most  electro-positive  binary,  the  base,  first.  If  the 
salt"  is  composed  of  more  than  one  equivalent  of  either  base  or 
acid,  then  the  number  of  equivalents  must  be  indicated  by 
2* 


18  CHEMICAL    SYMBOLS. 

coefficients.  The  ternary  symbol  thus  obtained  always  stands 
for  one  equivalent  of  the  compound,  and  the  weight  of  this 
equivalent  is  evidently  the  sum  of  the  weights  of  the  equiva- 
lents of  the  elements  entering  into  it. 

1  +  8  +  16  +  24  =  49. 

H  O,  S  O3  stands  for  1  equiv.  or  49  parts  of  Sulphate  of  Water 
(common  Sulphuric  Acid). 

1  +  8  +  14  4-  40  =  63. 

H  O,  N  O5  stands  for  1  equiv.  or  63  parts  of  Nitrate  of  Water 
(common  Nitric  Acid). 

39  +  8  +  6  +  16  =  69. 

K  O,  C  O2  stands  for  1  equiv.  or   69  parts  of  Carbonate  of 
Potassa. 

23  +  8  +  16  -(-  24  =  71. 

Na  0,  S  O3  stands  for  1  equiv.  or  71  parts  of  Sulphate  of  Soda. 

39  +  8  +  27  +  24  =  98. 

K  O,  Cr  O3  stands  for  1  equiv.  or  98  parts  of  Neutral  Chromate 
of  Potassa. 

39  +  8  +  2  (27  +  24)  =  149. 

K  0, 2  Cr  O3  stands  for  1  equiv.  or  149  parts  of  Bichromate  of 
Potassa. 

39  +  8  +  3  (27  +  24)  =  200. 

K  0, 3  Cr  O3  stands  for  1  equiv.  or  200  parts  of  Trichromate  of 
Potassa. 

104  +  8  4-  27  4-  24  =  163. 

PbO,  CrOs  stands  for  1  equiv.  or  163  parts  of  Neutral  Chro- 
mate of  Lead. 

2  (104  +  8)  +  27  4-  24  =  275. 

2  PbO,  Cr03  stands  for  1  equiv.  or  275  parts  of  Basic  Chromate 
of  Lead. 

In  order  to  express  two  or  more  equivalents  of  a  ternary, 
we  enclose  the  symbol  in  parentheses,  and  place  before  the 
whole  the  required  figure.     Thus, 


CHEMICAL    SYMBOLS.  19 

3  (Na  0,  S  Og)  stands  for  three  equivalents  of  Sulphate  of  Soda. 
5  (2  Pb  O,  Cr  O3)  stands  for  five  equivalents  of  Basic  Chromate 
of  Lead. 

Neutral  Salts.  —  The  larger  number  of  inorganic  acids  com- 
bine most  readily  with  one  equivalent  of  base,  and  the  salts  so 
formed  will  be  called  neutral  salts.  If  the  salts  contain  more 
equivalents  of  acid  or  base  than  one,  they  are  called  acid  or 
basic  salts  respectively.  There  are,  however,  some  acids  which 
combine  most  readily  with  two  or  three  equivalents  of  base,  in 
the  same  way  that  the  others  combine  with  one.  Such  acids 
are  called  bibasic  or  tribasic  acids,  in  order  to  distinguish  them 
from  the  rest,  which  are  frequently  called  monobasic.  Of 
bibasic  and  tribasic  acids,  the  most  important  in  inorganic 
chemistry  is  Phosphoric  Acid.  This  Is  known  in  three  differ- 
ent conditions.  In  the  first  of  these  it  is  monobasic,  in  the 
second  bibasic,  and  in  the  third  tribasic,  the  last  being  the 
ordinary  condition.  The  three  conditions  are  designated  by 
the  symbol  aP  O5  ;  ^P  O5 ;  ^P  O5.  The  acid  eP  O5  forms  neutral 
salts  when  combined  with  three  equivalents  of  base,  the  acid 
bP  Og  when  combined  with  two,  and  the  acid  ^P  O5  when  com- 
bined with  only  one.  There  are  three  compounds  of  the  acid 
and  water  corresponding  to  the  three  conditions,  which  are  rep- 
resented in  symbols  by  H  O,  ^P  Oj ;  2  H  0,  tP  Og ;  3  H  O,  ,P  O5. 
In  these  compounds  we  can  substitute  for  the  equivalents  of 
water  equivalents  of  other  bases,  either  in  whole  or  in  part, 
forming  such  compounds  as  Na  O,  ^P  Og  ;  2  Na  O,  bP  Og ; 
3]S-aO,,P05  ;  [HO,  2  NaO]  ^P  O5 ;  [2  H 0, Na 0]  ^P Og. 
The  equivalents  of  water  may  even  be  replaced  by  different 
bases,  as  in  the  compounds  [Na  O,  Pb  0]  bP  Og  ;  [H  0,  Na  0, 


20  CHEMICAL    STJIBOLS. 

K  0],  eP  O5  ;  [K  0,  2  Mg  0]  ,T  O5.    All  the  above  are  sym- 
bols of  neutral  salts. 

As  protoxide  bases  combine  most  readily  with  one  equivalent 
of  acid,  so  sesquioxide  bases  combine  most  readily  mth  three 
equivalents.  A  neutral  salt  of  a  sesquioxide  base  is  therefore 
one  which  contains  for  every  equivalent  of  base  three  equiva- 
lents of  a  monobasic  acid,  or  one  equivalent  of  a  tribasic  acid, 
and  for  every  two  equivalents  of  base  three  equivalents  of  a 
bibasic  acid.  Hence,  Fca  O3 ,  3  S  O3  ;  2  Fcg  O3  ,  3  tP  O5 ; 
P62  O3 ,  cP  O5  are  all  symbols  of  neutral  salts.  On  the  other 
hand,  Fea  O3 ,  S  O3 ;  4  AI2  O3 ,  3  ,P  O5 ;  2  Al^  O3 ,  cP  O5  are  sym- 
bols of  basic  salts.  It  will  be  noticed,  on  examining  the  above 
symbols,  that  neutral  salts  of  monobasic  acids  contain  as  many 
equivalents  of  acid  as  there  are  equivalents  of  oxygen  in  the 
base,  and  neutral  salts  of  bibasic  and  tribasic  acids  one  half 
and  one  third  as  many,  respectively.  This  rule  must  be  kept 
in  mind  when  writing  the  symbols  of  salts. 

Compound  Radicals.  —  There  is  a  large  class  of  substances 
which,  although  compound,  nevertheless  act  in  chemical  changes 
exactly  as  if  they  were  simple,  frequently  replacing  the  ele- 
ments themselves.  Such  substances  are  termed  compound 
radicals.  Many  of  these  radicals  have,  like  the  elements, 
received  arbitrary  names,  such  as  Cyanogen,  Ammonium, 
Ethyle,  Acetyle,  &c. ;  and,  moreover,  the  first  letter  or  letters 
of  these  names  are  frequently  used  as  their  symbols.  It  is 
best,  however,  to  write  out  the  symbols  of  the  elements  form- 
ing these  compounds,  and  enclose  the  whole  in  brackets  ;  thus, 
[N  H4]  stands  for  Ammonium,  [Cj  N]  for  Cyanogen,  [C4  H5] 
for  Ethyle,  [C4  H3]  for  Acetyle.     The  oxides  of  the  compound 


CHEMICAL    SYMBOLS.  21 

radicals,  like  those  of  the  elements,  may  be  divided  into  acids 
and  bases,  and  their  symbols  are  Avritten  exactly  like  those  of 
other  binaries.     Thus, 

[C4  H3]  O3  is  the  symbol  of  Acetic  Acid. 

[N  H4]  O  "  "  Oxide  of  Ammonium. 

[C4  H5]  0        "  "         Oxide  of  Ethyle. 

So,  also,  with  the  salts. 

[N  H4]  O,  [C4  H3]  O3  is  the  symbol  of  Acetate  of  Ammonia. 

[C4  Hs]  0,  [C4  H3]  O3        "        "      Acetate  ofOxide  of  Ethyle. 

Water  of  Crystallization.  —  Besides  the  water  of  constitution, 
which  frequently  forms  a  part  or  the  whole  of  the  base  of  a 
salt,  most  salts  combine  with  water  as  a  whole.  This  water  is 
held  in  combination  by  a  comparatively  feeble  affinity,  and 
may  be  generally  driven  off  by  exposing  the  salt  to  the  tem- 
perature of  100°  C,  and  sometimes  escapes  at  the  ordinary 
temperature  of  the  air,  the  crystals  of  the  salt  in  all  cases 
fallmg  into  powder.  Its  presence  is  essential  to  the  crystalhne 
condition  of  many  salts,  and  hence  the  name  "Water  of  Crys- 
tallization. The  presence  of  water  of  crystallization  in  a  salt 
is  expressed  in  symbols,  by  writing  after  the  symbol  of  the 
salt,  and  separated  from  it  by  a  period,  the  number  of  equiva- 
lents of  water.     Thus, 

Fe  O,  S  O3  .  7  H  0  is  the  symbol  of  Crystallized  Sulphate  of 

the  Oxide  of  Iron  (Green  Vitriol). 
H  0,  2  Na  O,  eP  O5  .  24  H  O  is  the  symbol  of  Crystallized 

Phosphate  of  Soda. 

The  same  salt,  when  crystallized  at  different  temperatures,  not 

unfrequently  combines  with  different  amounts  of  water  of  crys- 

,  tallization,  the  less  amounts  corresponding  to  the  higher  tem- 


22  CHEMICAL    SraiBOLS. 

peratures.  Thus,  the  Sulphate  of  Manganese  may  be  crystal- 
lized with  three  different  amounts  of  water  of  crystallization. 

Mn  O,  S  O3 .  7  H  O  when  crystallized  below  6°  Centigrade. 
Mn  0,  S  O3 .  5  H  O         "         «         between  7°  and  20°. 
Mn  O,  S  O3 .  4  H  O        «         «         between  20°  and  30°. 

The  crystalline  forms  of  these  three  compounds  are  entirely 
different  from  each  other,  proving  that  the  form  depends,  in 
part  at  least,  on  the  amount  of  water  which  the  salt  contains. 

The  symbols  of  other  ternary  compounds  are  written  like 
those  of  the  oxygen  salts,  and  therefore  require  no  further 
explanation.  Below  are  a  few  of  these  symbols,  together  with 
those  of  the  corresponding  oxygen  salts,  which  may  serve  as 
examples. 

K  O,  C  O2  =  Carbonate  of  Oxide  of  Potassium. 

K  S,  C  S2  =  Sulphocarbonate  of  Sulphide  of  Potassium. 

K  O,  As  O3  =  Arsenite  of  Oxide  of  Potassium. 

K  S,  As  S3  =  Sulphoarsenite  of  Sulphide  of  Potassium. 

3  Na  CI,  Sb  CI3  =  Double  Chloride  of  Antimony  and  Sodium. 
[NH4]  CljPtCla  =  Double  Chloride  of  Platinum  and  Potassium. 
K  I,  Pt  I2  =  Double  Iodide  of  Platinum  and  Potassium. 

Quaternaries.  —  The  symbols  of  the  double  salts  are  formed 
by  writing  together  the  symbols  of  the  two  salts  of  which  they 
consist,  separated  by  a  period.     Thus, 

K  O,  S  O3 .  iMg  O,  S  O3  .  6  H  O  =  Double  Sulphate  of  Mag- 
nesia and  Potassa. 

K  O,  S  O3 .  AI2  O3, 3  S  O3 .  24  H  0  ==  Double  Sulphate  of  Alu- 
mina and  Potassa  (Alum). 

When  the  salts  contain  water  of  crystallization,  the  amount  of 
this  water  expressed  in  equivalents  is  written  after  the  symbol 
of  the  salt,  as  already  explained. 


CHEMICAL    EEACTIONS. 


The  various  chemical  changes  to  which  all  matter  is  more 
or  less  liable  are  termed,  in  the  language  of  chemistiy,  reac- 
tions, and  the  agents  which  cause  these  changes,  reagents.  In 
every  chemical  reaction  we  must  distinguish  between  the  sub- 
stances which  are  involved  in  the  change  and  those  which  are 
produced  by  it.  The  first  will  be  termed  the  factors,  and  the 
last  the  products,  of  the  reaction.  As  matter  is  indestructible, 
it  follows  that  The  sum  of  the  weights  of  the  products  of  any 
reaction  must  always  be  equal  to  the  sum  of  the  weights  of  the 
factors.  This  statement  seems  at  first  sight  to  be  contradicted  by 
experience,  since  wood  and  many  other  combustible  substances 
are  apparently  consumed  by  burning.  In  all  such  cases,  how- 
ever, the  apparent  annihilation  of  the  substance  arises  from 
the  fact  that  the  products  of  the  change  are  invisible  gases ; 
and  when  these  are  collected,  their  weight  is  found  to  be  equal, 
not  only  to  that  of  the  substance,  but  also,  in  addition,  to  the 
weight  of  the  oxygen  from  the  air  consumed  in  the  process. 
As  the  products  and  factors  of  every  chemical  change  must  be 
equal,  it  follows  that  A  chemical  reaction  may  always  be  repre- 


24  CHEMICAL   REACTIONS. 

sented  in  an  equation  hy  writing  the  symbols  of  the  factors  in 
the  first  member,  and  those  of  the  products  in  the  second.  The 
I'eaction  of  sulphuric  acid  on  common  salt  may  be  represented 
by  the  following  equation : 

23  +  35    +     1+8  +  16+24    =     23  +  8  +  16+24     +     1  +  35  =  107. 

NaCl  +  ^0,  SOs  =  Mi  0,SOs  +HC1. 

The  correctness  of  this  may  be  proved  by  adding  together  the 
equivalents  of  both  sides,  when  the  sums  will  be  found  to  be 
equal.  In  like  manner,  the  reaction  of  a  solution  of  common 
phosphate  of  soda  on  a  solution  of  chloride  of  calcium  may  be 
represented  by  the  equation 

1+8  +  2  (23  +  8)+ 32+40    +    3(20+35)  = 

IIO,2I^aO,  ,FOs-^SCaCl-\-Aq*  = 

3  (20  +  8)  +  32+40    +    2(23  +  35)   +    1  +  85  =  308. 

3CaO,  eP05  +  2iVa  CI -\- JI CI -\- Aq. 

So,  also,  the  reaction  of  hydrochloric  acid  on  chalk,  which  may 
be  proved  like  the  other  two : 

CaO,CO^-{- !£  CI -\-Aq=  CaCl-]-II0-{-Aq-\-COa. 

Although  the  equation  is  the  most  concise,  and  therefore 
in  most  cases  the  best  form  of  representing  chemical  reac- 
tions, it  is  nevertheless  frequently  advantageous,  in  studying 
comphcated  changes,  to  adopt  a  more  graphic  method,  by 
which  the  various  steps  of  the  process  may  be  indicated.  The 
reactions  represented  by  the  preceding  equations  may  be  writ- 
ten thus  :  — 


*  The  symbol  Aq,  for  Aqua,  merely  indicates  the  condition  of  solution,  and 
is  not  to  be  regarded  in  adding  up  the  equivalents  in  order  to  prove  the  equa- 
tion. 


CHEMICAL   REACTIONS. 


25 


(2.) 


HO,2XaO,„  PO^ 


Chemical  reactions  may  be  classed  under  three  divisions. 

First,  those  reactions  in  which  a  compound  is  decomposed, 
and  divides  into  simpler  compounds  or  into  elements.  E.  g. 
when  oxide  of  mercury  is  heated,  it  is  decomposed  into  oxygen 
gas  and  metallic  mercury.     Thus, 

Hg0  =  ^y  +  O. 

Again,  when  Chlorate  of  Potassa  is  heated,  it  is  resolved  into 
oxygen  gas  and  chloride  of  potassium.     Thus, 

K  O,  CI  O5  ==  K  CI  +  6  O. 
3 


26  CHEMICAL   REACTIONS. 

So,  also,  when  sulphate  of  lead  is  heated,  it  is  resolved  into 
anhydrous  sulphuric  acid  and  oxide  of  lead.     Thus, 

PbO,S03=  PbO+SOs. 

Such  reactions  as  these  Avill  be  called  analytical,  and  the 
process  analysis. 

Second,  those  reactions  in  which  the  elements  are  united  to 
form  compounds,  or  compounds  of  a  lower  order  to  form  those 
of  a  higher.  E.  g.  when  hydrogen  and  carbon  bum  in  the  air, 
they  combine  with  oxygen  to  form  water  or  carbonic  acid. 
Thus, 

0  +  0=HO;  C  +  02=C02. 

Again,  when  anhydrous  sulphuric  acid  combines  with  lime  to 
form  sulphate  of  lime.     Thus, 

CaO  +  S03=CaO,S03. 

Reactions  like  these  will  be  called  sy7ithetical,  and  the  process 
synthesis. 

Third,  those  reactions  in  which  one  element  displaces  an- 
other. E.  G.  when  sodium  takes  the  place  of  hydrogen  in 
water,  or  zinc  the  place  of  hydrogen  in  dilute  sulphuric  acid. 
Thus, 

^6>  +  Na  =  iVa(9  +  H. 
Zn-^HO,SO^-\-Aq=ZHO,SO,-\-Aq^U.. 

This  division  includes  also  those  reactions  in  which  there  is 
a  mutual  interchange  of  elements  between  two  compounds. 
E.  g.  when  a  solution  of  chloride  of  barium  is  added  to  a 
solution  of  sulphate  of  soda,  the  sodium  and  barium  change 
places,  and  we  have  formed  an  insoluble  precipitate  of  sulphate 


CHEMICAL   REACTIONS. 


27 


of  baryta  and  chloride  of  sodium  (common  salt),  which  remains 
in  solution.     Thus, 

Ba  a-\-Na  0,  S 0^-^-  Aq  =  Bs, 0,^0, -\-  Fa  Cl+Aq. 

Reactions  like  these  will  be  called  metathetical,  and  the  process 
metathesis* 

Of  the  three  classes  of  chemical  reactions,  the  last  is  by  far 
the  most  important ;  indeed,  the  larger  number  of  reactions 
described  in  an  elementary  treatise  on  chemistry  are  examples 
of  metathesis.  All  metathetical  reactions  can  be  illustrated 
very  elegantly  with  the  aid  of  mechanical  diagrams,  as  fol- 
lows :  — 

(4)  (5.) 


They  are  easily  made  by  printing  with  stencils  on  the  larger 
piece  of  pasteboard,  A  B,  Fig.  4,  the  symbols  of  the  elements 
not  disturbed  in  the  reaction,  and  on  the  smaller  piece,  a  b,  the 
symbols  of  the  elements  which  exchange  places.  The  smaller 
piece  having  been  fastened  to  the  larger  by  means  of  an  eye- 
let at  O,  the  reaction  is  represented  by  merely  turning  it 
half  round.  (See  Fig.  5.)  If  the  symbols  of  the  interchang- 
ing elements  are  not  symmetrical  on  all  sides,  it  is  of  course 
necessary  to  make  them  reversible,  by  printing  each  on  a  sep- 
arate small  square  of  pasteboard,  fastened  by  an  eyelet  to  the 


*  From  the  Greek  /aerar I'^Tj/xt,  to  displace  or  to  transpose. 


28  CHEMICAL    REACTIONS. 

top  or  bottom  of  the  revolving  piece  a  b,  since  otherwise  the 
letters  would  be  inverted  when  the  diagram  is  turned.  This 
method  of  illustration  may,  with  a  little  ingenuity,  be  extended 
to  some  of  the  most  complicated  cases  of  chemical  change. 

The  most  important  condition  of  chemical  action  is,  that  the 
particles  of  the  substances  involved  in  the  change  should  be 
indued  with  freedom  of  motion.  This  condition  is  generally 
fulfilled,  both  in  nature  and  in  our  laboratories,  by  bringing  the 
substances  together  in  solution,  either  in  water  or  in  some  other 
fluid.  When  substances  are  brought  together  in  solution,  there 
are  two  circumstances  which,  more  than  any  others,  determine 
the  nature  and  extent  of  the  resulting  change. 

First,  If  hy  an  interchange  of  analogous  elements  an  insoluble 
coynpound  may  he  formed,  this  compound  always  separates  from 
the  fluid  as  a  precipitate.  As  this  circumstance  is  by  far  the 
most  important  of  all  in  determining  chemical  reactions,  it 
requires  full  illustration. 


Ba  0,  N  0: 
HO,  SO, 


1     ,     .         HO,NOs\    . 
|+^^=Ba6,S0:|-^?- 


In  these  examples,  and  in  general  throughout  the  volume,  the 
symbols  of  the  substances,  when  in  solution,  are  printed  in 
italic  letters,  and  the  solid  precipitate  in  Roman  letters.  The 
symbol  Aq,  as  already  stated,  stands  for  an  indefinite  amount 
of  water,  in  which  the  substances  are  supposed  to  be  dissolved. 
It  is  obvious,  from  the  above  examples,  that,  in  order  to  ascer- 


CHEMICAL   REACTIONS.  29 

tain  whether  two  salts  will  react  on  each  other,  when  brought 
together  in  solution,  so  as  to  form  a  precipitate,  it  is  only  neces- 
sary to  Avrite  the  symbol  of  one  under  that  of  the  other,  and 
interchange  the  symbols  of  the  metallic  elements.  If  either 
compound  whose  symbols  are  thus  formed  is  insoluble  in  the 
menstruum  present,  a  reaction  will  take  place,  and  the  insoluble 
compound  will  be  precipitated.  At  the  end  of  the  volume  will 
be  found  a  table,  reprinted  from  the  English  edition  of  Fre- 
senius's  Qualitative  Analysis,  by  means  of  which  the  student 
can  easily  ascertain  from  inspection  the  solubility  of  any  of 
the  more  frequently  occurring  binary  compounds  or  salts,  and 
thus  will  be  able  to  solve  the  following  problems. 

Problem  1.  If  chloride  of  barium  and  sulphate  of  soda  are 
mixed  together  in  solution,  wiU  there  be  a  reaction ;  and  if  so, 
what  will  be  formed  ? 

Problem  2.  If  chloride  of  sodium  and  nitrate  of  silver  are 
mixed  together  in  solution,  will  there  be  a  reaction,  &c.  ? 

Problem  3.  If  sulphide  of  hydrogen  and  nitrate  of  lead  are 
mixed  together  in  solution,  will  there  be  a  reaction  ? 

Problem  4.  If  sulphide  of  hydrogen  and  sulphate  of  zinc 
are  mixed  together  in  solution,  will  there  be  a  reaction  ? 

In  solving  the  last  problem,  it  must  be  noticed  that  the  fluid 
which  would  result  from  a  reaction  would  be  a  weak  acid,  in 
which  many  substances  are  soluble  which  would  be  insoluble 
in  pure  water,  as  may  be  seen  from  the  table. 

Problem  5.  If  chloride  of  sodium  and  sulphate  of  copper 
are  mixed  together  in  solution,  will  there  be  a  reaction  ? 

Problem  6.  If  sulphuric  acid  and  borate  of  soda  are  mixed 
together  in  solution,  will  there  be  a  reaction  ? 

It  will  be  found  that,  by  an  interchange  of  metallic  elements 
3* 


30  CHEMICAL    REACTIONS. 

in  the  last  two  examples,  no  insoluble  compound  will  be  formed, 
and  hence  the  conclusion  follows  from  our  data,  that  there  will 
be  no  precipitate.  We  must  not,  however,  conclude  from  this 
that  there  will  be  no  reaction,  since,  as  can  easily  be  seen,  it 
does  not  necessarily  follow,  because  the  possible  formation  of  an 
insoluble  compound  always  determines  a  reaction,  that  the  re- 
verse is  equally  true,  and  that  no  reaction  can  take  place  unless 
an  insoluble  compound  is  formed.  Indeed,  in  the  last  two  ex- 
amples, we  are  able,  from  incidental  phenomena,  to  determine 
satisfactorily  that  a  change  does  result ;  thus,  in  Problem  5, 
when  the  solutions  are  mixed,  the  blue  color  of  sulphate  of 
copper  changes  into  the  green  color  of  chloride  of  copper ; 
and  in  Problem  6,  if  sulphuric  acid  is  not  added  in  excess,  the 
claret  color  to  which  blue  litmus-paper  turns  in  the  mixed 
solution  proves  that  it  is  boracic  acid,  and  not  sulphuric  acid, 
wlaich  is  in  a  free  state  ;  nevertheless,  in  most  similar  cases  it 
is  impossible  to  determine,  unless  an  insoluble  compound  is 
formed,  whether  any  reaction  has  taken  place. 

Second.  The  circumstance  which,  next  to  insolubility,  is 
most  important  in  determining  metathetical  reactions,  is  vola- 
tility, and  it  may  be  laid  down  as  a  general  principle,  that,  If 
hy  an  interchange  of  analogous  elements  between  two  substances 
in  solution,  a  substance  can  be  formed,  which  is  volatile  at  the 
temperature  at  which  the  experiment  is  conducted,  such  an  inter- 
change cdioays  takes  place,  and  the  volatile  product  is  set  free. 
In  order  to  illustrate  this  principle,  a  few  examples  may  be 
adduced. 

1.  If  diluted  sulphuric  acid  is  poured  upon  granulated  zinc, 
a  brisk  evolution  of  hydrogen  gas  ensues,  and  sulphate  of  oxide 
of  zinc  is  retained  in  solution.     Thus, 


CHEMICAL    REACTIONS.  31 

Zn  1         jH 

HO,SO^-{-Aq\  —  \ZnO,SO,-\-Aq. 

In  this  example,  and  those  that  follow,  the  volatile  or  gaseous 
products  are  always  printed  with  a  full-face  type. 

2.  If  diluted  sulphuric  acid  is  poured  upon  protosulphide  of 
iron,  sulphide  of  hydrogen  gas  escapes,  and  sulphate  of  protox- 
ide of  iron  remains  in  solution.     Thus, 

FeS  )         (HS 

HO,SO^-{-Aq]  —  \Fe  0,SO^-\-Aq. 

It  will  be  noticed  from  the  last  two  reactions,  that  it  is  not 
essential  that  more  than  one  of  the  factors  of  the  reaction 
should  be  fluid,  or  in  solution. 

8.  If  strong  sulphuric  acid  is  poured  upon  common  salt,  and 
the  mixture  slightly  heated,  chlorohydric  acid  gas  is  evolved, 
and  bisulphate  of  soda  remains  dissolved  in  the  excess  of  sul- 
phuric acid.     Thus, 

Na  a  7         f  H  CI 


}  = 


HO,SOs .  HO,  ^6>3 1  —  XNa  0,  S  O3 .  HO,  S  0,. 

4.  If  strong  sulphuric  acid  is  poured  upon  nitre,  and  the 
temperature  of  the  mixture  slightly  elevated,  the  vapor  of 
nitric  acid  is  given  off,  and  bisulphate  of  potassa  is  formed. 
Thus, 

KO,NO,  \         fH0,JV05 

HO,  8  0^.H0,S0^\  —  \K0,  S  O3  .  HO,  S  O3. 

5.  If  diluted  nitric  acid  is  poured  upon  chalk,  or  any  other 
analogous  carbonate,  carbonic  acid  gas  is  set  free,  and  a  salt 
of  nitric  acid  formed.     Thus, 

CaO,  CO,  )  _  j  H0-\-CO, 

HO,NO,^Aq\  —  \  OaO,J^Os-\-Aq. 


32  CHEMICAL   KEACTIONS. 

It  is  very  frequently  the  case  that  two,  or  even  all  of  the 
three,  classes  of  chemical  reactions  are  combined,  and  going  on 
simultaneously,  in  a  single  experiment.  In  the  last  example, 
for  instance,  the  metathesis  is  succeeded  by  an  analysis  of  one 
of  the  products,  owing  to  the  want  of  aflBnity  between  C  O2  and 
H  O.  Again,  the  reaction  of  nitric  acid  on  copper  is  an  ex- 
ample where  all  three  varieties  of  reactions  are  combined. 
Three  equivalents  of  copper  react  on  four  equivalents  of  nitric 
acid,  and  the  reaction  may  be  conveniently  studied  in  two  parts. 
In  the  first  part,  the  copper  is  oxidized  by  one  of  the  equiva- 
lents of  the  acid ;  here  Ave  have  analysis  accompanied  by  syn- 
thesis ;  and  in  the  second  part  the  copper  changes  place  with 
the  hydrogen  of  the  acid,  a  case  of  metathesis. 

1st.    3Cu  +  ^0,  iV^(95  =  3(CuO,  H0)-f-]\O2. 

HO,NO,\,  SGuO,NOs    1    ,      . 

Z"^-    CuO,  H0[  +^^~  l^O  +  ^Oj  +^^- 

The  whole  reaction  combined  may  be  expressed  thus : 

3  Cu  +  4  {HO,  NO,)  -{-Aq  =  3(at  0,N0,)  +^g  +  JVO^. 

Such  mixed  processes  are  very  common  in  all  complex  cases 
of  chemical  change. 

StocJtiometrieal  Problems.  —  The  chemical  symbols  enable 
us  not  only  to  represent  chemical  changes,  but  also  to  calculate 
exactly  the  amounts  of  the  substances  required  in  any  given 
process,  as  well  as  the  amounts  of  the  products  wliich  it  will 
yield.  The  method  of  making  such  calculation  can  best  be 
illustrated  by  examples.  In  these  examples  the  weights  and 
measures  will  be  given  according  to  the  French  decimal  system, 
which  is  now  very  generally  used  in  chemical  laboratories,  and 


CHEMICAL   REACTIONS.  33 

which,  on  account  of  the  very  simple  relation  between  the  units 
of  measure  and  of  weight,  greatly  facilitates  stochiometrical 
calculations.  The  French  measures  and  weights  can,  when 
required,  be  very  easily  reduced  to  the  English  standards,  by 
means  of  a  table  at  the  end  of  the  volume. 

Pi'oblem  1.  We  have  given  10  kilogrammes  of  common 
salt,  and  it  is  required  to  calculate  how  much  chlorohydric  acid 
gas  can  be  obtained  from  it  by  treating  with  sulphuric  acid. 
The  reaction  is 

23+35.5   =   58.5.  1+35.5  =  36.5. 

I^eiCl-{- ffO,S03-{-Aq  =  mO,SOs-\-Aq-\-U€L 

Hence  one  equivalent,  or  58.5  parts,  of  common  salt,  will  yield 
one  equivalent,  or  36.5  parts,  of  chlorohydric  acid  gas.  There- 
fore the  amount  which  10  kilogrammes  will  yield  can  be  cal- 
culated from  the  proportion 

53.5  36.5 

Na  CI :  H  CI  =  10  :  X  =  6.239  kilogrammes,  Ans. 

Problem  2.  It  is  required  to  calculate  how  much  chlorine 
gas  can  be  obtained  from  chlorohydi'ic  acid  with  6  grammes  of 
hyperoxide  of  manganese.  The  equation  representing  the 
reaction  is 

28.6+16  44.6.  35.5. 

Mn02  -\- 2  H  01 -\-  Aq  =  Ml  a  -\-  2  HOi^Aq  +  €1. 

Hence  the  amount  of  hyperoxide  of  manganese  represented  by 
Mn  O2,  or  44.6  parts,  yields  an  amount  of  chlorine  gas  repre- 
sented by  CI,  or  35.5  parts,  and  we  have  the  proportion, 

44.6      35.5 
Mn  O2 :  CI  =  6  :  a;  =  4.775  grammes. 

Problem  3.   It  is  required  to  calculate  how  much  sulphuric 


34  CHEBIICAL   REACTIONS. 

acid   and  nitre  must  be  used  to  make  250  gi'ammes  of  the 
strongest  nitric  acid. 

101.2  98  63 

K0,-^0,+2(JI0,S0,)=K0,S0s.II0,S0s-\-nO,NO,. 
Hence, 

63  98 

H0,I^0si2{II0,S  O3)  =  250  :  x. 

63  101.2 

HO,NO^ :  K 0,  N Og  =  250  :  x. 

From  the  above  examples  we  can  deduce  the  following  gen- 
eral rule  for  calculating  from  the  amount  of  any  given  factor 
of  a  chemical  reaction  the  amount  of  the  products,  or  the 
reverse.  Express  the  reaction  in  an  equation :  make  then  the 
proportion,  As  the  symbol  of  the  substance  given  is  to  the  sym- 
bol of  the  substance  required,  so  is  the  amount  of  the  substance 
given  to  x,  the  amount  of  the  substance  required ;  reduce  the 
symbols  to  numbers,  and  calcidate  the  value  of  x. 

On  account  of  the  very  great  lightness,  the  amount  of  a  gas 
is  very  much  more  frequently  estimated  by  measure  than  by 
weight.  At  the  end  of  the  volume  a  table  will  be  found  giving 
the  weight  in  grammes  of  one  thousand  cubic  centimetres,  or 
one  litre,  of  each  of  the  most  important  gases.  With  the  aid 
of  this  table  such  problems  as  the  following  may  be  solved. 

Problem  4.  How  much  chlorate  of  potassa  must  be  used  to 
obtain  one  litre  of  oxygen  gas?  One  litre  of  oxygen  gas 
weighs  1.43  grammes. 

K  0,  CI  O5  =  K  CI  +  6  O. 

48  122.7 

6  0   :   K  0,  CI  O5  =  1.43   :  x. 


CHEMICAL    REACTIONS.  35 

Problem  5.  How  much  zinc  and  sulphuric  acid  must  be  used 
to  obtain  4  litres  of  hydrogen  gas  ?  Four  litres  of  hydrogen 
weigh  0.357  grammes. 

Zn  -^HO,SOs-\-Aq=  Zn  0,  S  0^ -\- Aq -\-  H. 

1        32.6 

H  :  Zn  ==  0.357  :  a;  =  amount  of  zinc  required. 

1  49 

H  :  HO,  S03  =  0.357  :  x  =  amount  of  sulphuric  acid  required. 

Problem  6.  If  ten  grammes  of  water  are  decomposed  by 
galvanism,  how  large  a  volume  of  mixed  gases  will  they  give  ? 

^0  =  M+O. 

9  1 

JIO  :  H  =  10  :  x  =  i^-  grammes  of  hydrogen.^ 

9  8 

^0:0=10  :  x  =  ^^-  grammes  of  oxygen. 

J^  grammes  of  hydrogen  occupy  12.^429  cubic  centimetres. 
-^  grammes  of  oxygen  occupy         6.214  "         " 

The  mixed  gases  occupy  18,644  "         " 

And  hence  *  water,  when  decomposed  into  its  elements,  expands 
1864  times. 

The  use  of  chemical  symbols,  both  in  expressing  chemical 
reactions  and  in  stochiometrical  calculations,  having  been  ex- 
plained, they  will  be  used  in  the  following  pages  to  illustrate 
the  text  of  Stockhardt's  Elements  of  Chemistry.  The  reac- 
tions described  in  that  work  are  represented  in  the  form  of 
equations,  the  first  member  containing  always  the  factors,  and 

*  It  must  be  remembered  that  one  cubic  centimetre  of  water  weighs  one 
sramme. 


36  CHEMICAL   KE ACTIONS. 

the  second,  the  products  of  the  process.  It  remams  for  the 
student  to  work  out  the  reaction,  and  represent  it  in  the  manner 
explained  on  page  25.  The  equivalents  of  water  which  are  set 
free  or  formed  during  a  reaction,  are  not  generally  indicated, 
but  are  merged  in  the  general  symbol  ^5';  and  the  student 
will  frequently  be  obliged  to  supply  these  equivalents  in  order 
to  work  out  the  reaction.  The  symbols  of  solids  are  printed 
in  Roman  letters,  those  of  fluids  in  italics,  and  those  of  gases 
in  full-faced  type.  "When,  however,  the  solid  or  gas  is  dis- 
solved in  water,  the  symbol  is  printed  in  italics,  followed  by 
the  general  symbol  Aq,  or  aq.  Aq  always  stands  for  an  indefi- 
nite and  large  amount  of  water  ;  ag,  for  an  indefinite  but  small 
amount  of  water.  The  color  of  a  substance,  especially  of  pre- 
cipitates, is  frequently  printed  above  its  symbol.  The  head- 
ings and  figures,  or  letters  at  the  side  of  the  page,  refer  to  the 
sections  of  the  above-mentioned  book. 

In  order  still  further  to  illustrate  the  subject,  a  large  number 
of  problems  have  been  added  to  the  reactions,  which  the  stu- 
dent is  expected  to  solve.  The  method  of  solving  these  prob- 
lems can,  in  most  cases,  be  deduced  from  the  explanations 
already  given,  and  from  the  sections  of  the  text-book ;  in  all 
other  cases  the  method  is  explained  under  the  problem. 
Throughout  the  following  pages,  all  gases  and  vapors  are  sup- 
posed to  be  measured  at  the  temperature  of  0°  and  when  the 
barometer  stands  at  76  centimetres,  unless  some  other  tempera- 
ture or  barometric  pressure  is  expressly  stated.  All  specific 
gravities  are  referred  to  water  at  its  maximum  density  (at  4°). 
The  temperatures  are  all  given  on  the  Centigrade  scale,  and 
the  weights  and  measures  according  to  the  French  decimal 
system. 


EEACTIONS  AND  PROBLEMS. 


WEIGHING  AND   MEASUEING. 


10.     Problems  on  French  System  of  Weights  and  Measures. 

1.  Reduce  by  means  of  the  table  at  the  end  of  the  book,  — 
a.  30  inches  to  fractions  of  a  metre. 

h.  76  centimetres  to  English  inches. 

c.  36  feet  to  metres. 

d.  10  metres  to  feet  and  inches. 

2.  Reduce  by  means  of  the  table  at  the  end  of  the  book,  — 
a.  8  lbs.  6  oz.  to  grammes. 

h.  7640  grammes  to  English  apothecaries'  weight. 
c.  45  grains  to  grammes. 

3.  "What  is  the  diameter,  and  what  is  the  circumference,  of 
the  globe  in  French  measure  ? 

4.  What  is  the  distance  from  Dunkirk  in  France  to  Barce- 
lona in  Spain  ?  The  latitude  of  Dunkirk  =51°  3',  that  of 
Barcelona  =  41°  22',  and  the  two  places  are  on  the  same 
meridian. 

5.  "Were  our  globe  composed  entirely  of  water  at  its  great- 
est density,  what  would  be  its  weight  in  kilogrammes  ? 

6.  What  is  the  weight  of  one  cubic  decimetre  of  water  ? 

7.  Reduce  by  means  of  the  table  at  the  end  of  the  book,  — 
a.  4  pints  to  litres  and  cubic  centimetres. 

h.  5  gallons  to  litres  and  cubic  centimetres. 


40  WEIGHING   AND    MEASURING. 

c.  5  litres  to  English  measure. 

d.  4  cubic  centimetres  to  English  measure. 

11-17.  Problems  on  Specijlc  Gravity. 

1.  The  specific  gravity  of  iron  =  7.84.  What  is  the  weight 
of  1  cubic  centimetre,  4  cubic  centimetres,  &c.  of  the  metal  m 
grammes  ? 

2.  The  specific  gravity  of  alcohol  =  0.81.  "What  is  the 
weiglit  of  one  litre  in  grammes  ?  of  45  cubic  centimetres,  &c.  ? 

3.  The  specific  gravity  of  sulphuric  acid  =  1.85.  K  you 
wish  to  use  in  a  chemical  experiment  250  grammes,  how  much 
must  you  measure  out  ? 

4.  Knowing  the  specific  gravity  of  any  given  substance,  how 
can  you  calculate  the  weight  corresponding  to  any  given  meas- 
ure, or  the  measure  corresponding  to  any  given  weight  ?  Give 
a  general  algebraic  formula  for  the  purpose,  representing  spe- 
cific gravity,  weight,  and  volume,  by  Sp.  Gr.,  W.,  and  V. 

6.  Determine  the  Sp.  Gr.  of  absolute  alcohol  from  the 
following  data :  — 

Grammes. 

Weight  of  bottle  empty,  4.326 

"  "        fiUed  with  water  at  4°,  19.654 

«  «  "  alcohol  at  0°,  16.741 

6.  Determine  the  Sp.  Gr.  of  lead  shot  from  the  following 
data :  — 

Grammes. 

Weight  of  bottle  empty,  4.326 

"       "       «      filled  with  water  at  4°,  19.654 

«       "    shot,  15.456 

"       "    bottle,  shot,  and  water,  33.766 

7.  Determine  the  Sp.  Gr.  of  iron  from  the  following  data :  — 

Grammes. 

Weight  of  iron  in  air,  3.92 

"        "        «      water,  3.42 


WEIGHING   AND   MEASUKING.  41 

8.  Determine  the  Sp.  Gr.  of  copper  from 

Grammes. 

Weight  of  copper  in  air,  10.000 

«         "         «       water,  8.864 

9.  Determine  the  Sp.  Gr.  of  ash  wood  from 

Grammes. 

Weight  of  wood  in  air,  25.350 

"         a  copper  sinker,  11.000 

"         wood  and  sinker  under  water,  5.100 

10.  How  much  bulk  must  a  hollow  vessel  of  copper  fill, 
weighing  one  kilogramme,  which  will  just  float  in  water  ? 

11.  How  much  bulk  must  a  hollow  vessel  of  iron  occupy, 
weighing  ten  kilogrammes,  which  sinks  one  half  in  water  ? 

12.  An  alloy  of  gold  and  silver  weighs  ten  kilogrammes 
in  the  air,  and  9.735  kilogrammes  in  water.  What  are  the 
proportions  of  gold  and  silver  ?  Sp.  Gr.  of  gold  =  19.2,  of 
silver  =  10.5. 

Solution.  —  In  the  French  system  the  volume  of  a  solid  in  cubic  centimetres 

w 
equals  its  weight  in  grammes  divided  by  its  Sp.  Gr.,  or  V  =  g — q^.    Since  one 

cubic  centimetre  of  water  weighs  one  gramme,  the  volume  of  a  solid  in  cubic 

centimetres  is  equal  to  the  weight  of  water  it  displaces  in  grammes.   Hence  the 

w 
weight  of  water  displaced  =  g — ^.    Put,  then,  x  =  weight  of  gold  in  alloy, 

10  —  X  will  equal  weight  of  silver,    y^  =  weight  of  water  displaced  by  gold, 

and  -jjy^  =  weight  of  water  displaced  by  silver.    Hence  -^  +     ~^  =  0.265. 

13.  An  alloy  of  copper  and  silver  weighs  37  kilogrammes 
in  the  air,  and  loses  3.666  kilogrammes  when  weighed  in  water. 
What  are  the  proportions  of  silver  and  copper  ? 

22.  Problems  on  Expansion  of  Liquids  and  Gases. 

1.  If  34.562  cubic  centimetres  of  mercury  at  0°  are  heated 
to  100°,  what  increase  of  volume  do  they  undergo,  and  what 
is  the  increased  volume  ? 

Solution.  —  The  small  fraction  of  its  volume  by  which  one  c.  c.  of  a  liquid 

4* 


42  WEIGHING  AND   MEASURING. 

or  gas  increases  ■when  heated  from  0°  to  1°,  is  called  the  coefficient  ofeocpandon 
of  that  liquid  or  gas.  The  coefficient  of  expansion  of  mercury,  for  example, 
=  0.00018,  that  is,  one  c.  c.  of  mercury  at  0°  becomes  1.00018  c.  c.  at  1°.  If 
Tve  assume  that  the  expansion  is  proportional  to  the  temperature,  then  one 
c.  c.  at  Qo  becomes  1.00036  at  2°,  1.0009  at  5°,  and  1.018  at  100°.  Hence, 
84.562  c.  c.  of  mercury  would  become,  at  100°,  (34.562  X  1.018)  c.  c.  To  make 
this  solution  general,  let  h  =  coefficient  of  expansion ;  then  (1  +  ^)  =  increased 
volume  of  one  c.  c.  when  heated  from  0^  to  1°,  and  (1  +  <  h)  =  increased 
volume  of  one  c.  c.  when  heated  from  0°  to  <°,  and  V  (1  +  <  ^)  =  increased 
volume  of  V  c.  c.  when  heated  from  0°  to  t° ;  representing  by  V  the  increased 
volume,  we  have 

Y'=Y  {1  +  toli), 

from  which  the  increased  volume  of  any  liquid  or  gas  may  be  calculated  when 
the  volume  at  0°,  the  coefficient  of  expansion,  and  the  temperature  are  knoAvn. 
It  is  not  true,  as  we  have  assumed,  that  liquids  expand  twice  as  much  for  two 
degi'ees,  three  times  as  much  for  three  degrees,  &c.,  as  they  do  for  one;  but, 
on  the  contrary,  the  rate  of  expansion  slowly  increases  with  the  temperature. 
For  example,  one  c.  c.  of  mercury  at  0°  becomes  1.000179  at  1°,  but  one  c.  c. 
at  300o  becomes  1.000194  at  331°.  This  difference  of  rate,  however,  is  so 
small,  that  we  can  neglect  it,  except  in  the  most  refined  experiments,  more 
especially  if  we  use  not  the  coefficient  observed  at  any  particular  temperature, 
but  a  mean  coefficient  obtained  by  observing  the  total  amount  of  expansion 
between  0°  and  100°,  and  then  dividing  the  result  by  100,  by  which  we  averrge 
the  error.  Again,  experiments  on  the  expansion  of  fluids  are  commonly  con- 
ducted in  glass  vessels,  which  expand  themselves  by  heat,  and  therefore  cause 
the  expansion  to  appear  less  than  it  is.  If  they  expanded  as  much  as  the 
fluid,  it  is  evident  that  the  fluid  would  not  appear  to  expand  at  all.  They  in 
fact  expand  much  less  than  the  fluids,  but,  nevertheless,  sufficiently  to  make 
a  material  difference  between  the  absolute  expansion  of  a  fluid,  and  its  apparent 
expansion  in  glass  vessels.  The  mean  absolute  coefficient  of  mercury  between 
0°  and  100°  is  1.000181.  The  apparent  expansion  in  glass  between  0°  and 
100°  is  1.000156.  In  Table  IV.,  at  the  end  of  the  volume,  will  be  found  the  mean 
coefficients  of  expansion  in  glass  of  some  of  the  more  important  fluids.  The 
rate  of  expansion  of  water  varies  so  rapidly  and  so  anomalously,  that  no  use 
should  be  made  of  its  coefficient  except  in  experiments  extending  over  100°. 
When  it  is  required  to  determine  the  amount  of  expansion  between  naiTower 
limits,  use  should  be  made  of  Table  V.,  which  gives  the  volume  to  which  one 
cubic  centimetre  of  water  at  0°  or  at  4°  increases,  when  heated  to  the  tem- 
peratures at  the  side  of  the  table.  It  also  gives  the  Sp.  Gr.  of  water  at  different 
temperatures,  when  either  water  at  0°  or  at  4°  is  taken  as  the  unit. 

2.  What  will  be  the  volume  of  5.346  c.  c.  of  water  at  0°, 
when  heated  to  100°  ? 

3.  What  will  be  the  volume  of  250  c.  c.  of  oil  of  turpentine, 
at  0°,  when  heated  to  50°  ? 


WEIGHING   AND    MEASUKING.  43 

4.  What  will  be  the  volume  of  35  c.  c.  of  water  at  0°,  when 
heated  to  4°,  to  25°,  to  40°,  and  to  84°  ? 

17.  Problems  on  reducing  Specific  Gravities  to  the  Standard 
Temperature. 

The  specific  gravities  of  both  liquids  and  solids  are  supposed 
to  be  referred  to  water  at  4°,  its  greatest  density ;  but  in  prac- 
tice we  always  use  water  at  a  much  higher  temperature,  and  it 
becomes  therefore  necessary  to  reduce  the  results  to  4°,  or  in 
other  words,  to  calculate  what  would  have  been  the  result  had 
the  temperature  been  4°  during  the  experiment.  Hence  an 
important  class  of  problems  like  the  following :  — 

1.  The  Sp.  Gr.  of  zinc  was  found  to  be  7.1582  when  the 

temperature  of  the  water  was  15°.     What  would  have  been 

the  Sp.  Gr.  at  4°  ? 

Solution.  —  Sp.  Gr.  of  water  at  4°  :  Sp.  Gr.  at  15°  =  Sp.  Gr.  of  zinc  at 
15°  :  Sp.  Gr.  at  4°;      or      1  :  0.9992647  =  7.1582  :  x. 

2.  The  Sp.  Gr.  of  antimony  was  found  to  be  6.681  when 
the  temperature  of  the  water  was  15°.  What  would  have 
been  the  Sp.  Gr.  at  4°  ?  Ans.  6.677. 

3.  The  Sp.  Gr.  of  an  alloy  of  zinc  and  antimony  was  found 
from  the  following  data  :  — 

Grammes. 

Weight  of  the  alloy,  4.4106 

"         Sp.  Gr.  bottle,  9.0560 

"  "  "    full  of  water  at  4°,  19.0910 

«         bottle,  alloy,  and  water  at  14°.6,  22.8035 

Ans.  6.375. 

4.  Find  the  Sp.  Gr.  of  metallic  zinc  from  the  following 
data :  — 

Grammes. 

Weight  of  the  zinc,  12.4145 

«         bottle,  9.0560 

"  «       full  of  water  at  18°,  19.0790 

"  "       zinc  and  water  at  12°.4,  29.7663 

Ans.  7.153. 


44  WEIGHING   AND   MEASURING. 

24.  Problems  on  reducing  Centigrade  Degrees  to  Fahrenheit, 
and  the  reverse. 

1.  What  do  —40°,  —20°,  —15°,  —9°,  0°,  4°,  10°,  13°,  15°, 
80°,  100°,  150°  Cent,  correspond  to  on  the  Fahrenheit  scale  ? 

Ride.  —  Double  the  number  of  degrees,  subtract  one  tenth 
of  the  whole,  and  add  32  if  the  degrees  are  above  0°,  or  sub- 
tract 32  if  they  are  below. 

2.  What  do  —40°,  —32°,  —18°,  —7°,  0°,  10°,  32°,  42°,  50°, 
70°,  90°,  100°,  150°,  212°,  300°,  450°  Fahr.  correspond  to  on 
the  Centigrade  scale  ? 

Rule.  —  Subtract  32  if  above  0°,  or  add  32  if  below,  add 
one  ninth  to  the  result,  and  divide  by  two. 

27.  Problems  on  Expansion  of  Solids. 

1.  What  will  be  the  length  of  a  rod  of  iron  424.56  metres 
long  at  0°,  when  heated  to  20°  ? 

Solution.  —  The  small  fraction  of  its  length  by  which  a  rod  of  iron,  or  of  any 
other  solid,  one  metre  long,  expands  when  heated  from  0°  to  1°,  is  called  the 
coefficient  of  linear  expansion  of  the  solid.  A  bar  of  iron  one  metre  long 
at  0°  becomes  1.0000122  at  1°,  and  the  small  fraction  0.0000122  is  the 
coefficient  of  linear  expansion  of  iron.  Eepresenting  this  coefficient  by  h, 
Ave  have  for  the  new  length  of  the  rod,  at  1°,  V  =  1  (1  +  X"),  and  at  io^ 
V  =  l{\-\-t'k).  We  may  assume,  in  the  case  of  solids,  that  the  expansion  is 
proportional  to  the  temperature,  especially  if  we  deduce  our  coefficient  from 
experiments  made  between  0°  and  100°,  as  described  under  examples  on  the 
expansion  of  fluids.  Substituting  for  ?,  ^,  and  t  the  values  given  in  the  problem, 
we  have  11  =  424.56  (1  +  20  x  0.0000122).  The  coefficients  of  linear  expan- 
sion for  a  number  of  solids  are  given  in  Table  IV.  at  the  end  of  the  volume. 

2.  What  will  be  the  length  of  a  rod  of  copper  2.365  metres 
long,  at  0°,  when  heated  to  100°  ? 

3.  What  will  be  the  length  of  a  rod  of  silver  0.760  metres 
long,  at  12°,  when  heated  to  20°  ? 

Solution.  —  Denoting  by  I  the  unknown  length  of  the  rod  at  0°,  hy  I'  the 
known  length  at  i°,  and  I"  the  required  length  at  t'o,  we  have  as  above, 

l'=l{l  +  tk),    and    l"^Hl  +  fk). 


WEIGHING  AND    MEASURING.  45 

By  combining  these,  we  obtain 

^"  =  ^  (rrfi)  =  ^'  [1  +  ^  «'  -  0  +  &c.]. 

All  the  terms  of  the  quotient  may  be  neglected  after  the  first,  because  they 
contain  powers  of  the  already  very  small  fraction  Ic.  Substituting  the  values 
in  the  above  equation,  we  get 

l'  =z  0.760  [1  +  0.000019  (20  —  12)]. 

4.  One  of  the  large  iron  tubes  of  the  Britannia  Bridge  over 
the  Menai  Straits  is  143.253  metres  long.  What  increase  of 
length  does  it  undergo  between  8°  and  20°  Centigrade  ? 

5.  What  is  the  increased  capacity  of  a  globe  of  glass  which 
holds  exactly  one  litre,  at  0°,  when  heated  to  250°  ? 

Solution.  —  If  we  consider  for  a  moment  the  glass  globe  as  forming  the  out- 
side shell  of  a  solid  globe  of  glass,  it  is  evident  that  the  increased  capacity  of 
the  globe  will  be  equal  to  the  increased  volume  of  this  solid  globe,  which  we 
have  supposed  for  a  moment  to  fill  the  interior.  The  problem,  therefore, 
resolves  itself  into  calculating  the  amount  of  cubic  expansion  of  a  solid  glass 
globe  having  a  volume  equal  to  one  litre,  or  1000  c.  c.  The  coefficient  of  linear 
expansion  of  glass  is  given  in  the  table  as  0.00001.  The  coefficient  of  cubic 
expansion  is  always  three  times  *  as  great  as  the  linear  expansion  ;  in  the 
case  of  glass,  therefore,  it  is  0.00003.  By  this  is  meant  that,  if  a  rod  of  glass 
one  metre  long  at  0°  becomes  1.00001  long  at  1°,  then  a  cube  of  glass  of  one 
cubic  centimetre  at  0°  becomes  1.00003  c.  c.  at  1°.  Using,  then,  the  equa- 
tion 

we  obtain  by  substitution 

V  =  1000  (1  +  250  X  0.00003)  =  1007.5  c.  c. 

6.  What  is  the  increased  capacity  of  a  globe  of  glass  which 
holds  exactly  495  c.  c.  at  15°,  when  heated  to  300°  ? 


*  Each  edge  of  a  cube  of  glass  one  centimetre  long  at  0°,  would  become 
(1  +  h)  c.  m.  long  at  1°.  The  increased  volume  of  the  cube  would  be  equal  to 
(1  -)-  Z;)3  =  1  +  3  ^-  4-  3  X-2  +  7j3;  but  as  Ic  is  an  exceedingly  small  fraction, 
Z;2  and  Z;3  may  be  neglected  in  comparison,  so  that  a  cube  of  glass  of  one  c.  c, 
at  0°  becomes  {1  +  Zh)^i  1°,  which  proves  that  the  amount  of  cubic  expan- 
sion is  three  times  as  great  as  the  linear. 


NON-METALLIC  ELEMENTS,  or 
METALLOIDS. 


FIRST    GEOUP  :    ORGANOGENS. 

Oxygen  (0). 


56.  Hg  b  =  ^  +  ©. 

1.  How  much  oxygen  can  be  obtained  by  heating  108 
grammes,  250  grammes,  25  grammes,  or  5  grammes,  of 
red  oxide  of  mercury  ? 

2.  How  many  cubic  centimetres  of  oxygen  can  be 
obtained  from  the  amounts  of  oxide  of  mercury  given 
in  the  last  example  ? 

3.  How  much  oxide  of  mercury  would  be  required  to 
yield  one  litre  of  oxygen  by  the  process  of  1  ? 

4.  How  much  mercury  would  remain  after  the  experi- 
ment of  last  example  ? 

59.  KO,  CIO5  =  KCl+6©. 

1.  How  much  oxygen  can  be  obtained  from  1  kilo- 
gi-amme,  50  grammes,  or  5  grammes,  of  chlorate  of  po- 
tassa  ? 

2.  How  much  chloride  of  potassium  would  remain  after 
the  oxygen  in  the  last  examples  had  been  driven  off? 


NON-METALLIC   ELEMENTS,    OR    METALLOIDS.  47 

3.  How  much  cMorate  of  potassa  would  be  required  to 
make  one  litre  of  oxygen  ? 

63.  C  +  2  O  =  C  O2. 

1.  How  much  carbonic  acid  gas  will  be  formed  by  burn- 
ing 5  grammes  of  carbon  ? 

2.  How  many  cubic  centimetres  of  carbonic  acid  will 
be  formed  by  burning  5  grammes  of  carbon  ? 

3.  How  much  oxygen  will  be  consumed  in  the  last  two 
examples  ? 

4.  How  many  cubic  centimetres  of  carbonic  acid  will 
be  formed  from  one  litre  of  oxygen,  by  burning  in  it 
carbon  ? 

64.  S  +  2  O  =  S  O2. 

1.  How  much  sulphurous  acid  gas  wiU  be  formed  by 
burning  10  grammes  of  sulphur  ? 

2.  How  much  oxygen  will  be  consumed  in  the  last 
example  ? 

3.  How  many  cubic  centimetres  of  sulphurous  acid  wiU 
be  formed  by  burning  sulphur  in  one  litre  of  oxygen  ? 

4.  Assuming  that  one  litre  of  oxygen  yields  exactly  one 
litre  of  sulphurous  acid,  what  is  the  Sp.  Gr.  of  S  Og  gas  ? 

65.  P  +  50  =  P05. 

1.  How  much  phosphoric  acid  can  be  formed  from  48 
grammes  of  phosphorus  ? 

2.  How  much  phosphorus  will  exactly  consume  one 
litre  of  oxygen  ? 

3.  How  much  phosphorus  is  required  in  order  to  make 
250  grammes  of  phosphoric  acid  ? 


48  NON-METALLIC   ELEMENTS,    OR   METALLOIDS. 

67.  NaH-©  =  NaO. 

1.  How  much  oxide  of  sodium  can  be  made  from 
28.75  grammes  of  sodium  ? 

2.  How  much  oxide  of  sodium  can  be  made  with  one 
litre  of  oxygen  ?  and  how  mucli  sodium  will  be  consumed 
in  the  experiment  ? 

68.  3  Fe  +  4  O  =  Fe  O,  Fea  O3. 

71.  2NaO-^POs-^Aq=  HO,  2NaO,PO,-{-  Aq. 

The  ciystallized  salt  =  [HO,  2NaO]  PO5  .  24HO. 

72.  Ca0^aO^^Aq=  CaO,  CO2  +  Aq. 

73.  CaO-{-  SOi-\-Aq=  Ca  0,S0^-\-  Aq 

1.  How  much  lime  would  be  required  to  neutralize  20 
grammes  of  sulphurous  acid  ? 

2.  How  much  lime  would  be  required  to  neutralize 
the  sulphurous  acid  obtained  by  burning  5  grammes  of 
sulphur  ? 

3.  How  much  lime  would  be  required  to  neutrahze  one 
litre  of  sulphurous  acid  gas  ? 

79.  3  Mn  Oi  =  Mn  O,  Mng  O3  +  2  O. 

1.  How  much  oxygen  can  be  obtained  from  one  kilo- 
gramme of  hyperoxide  of  manganese  by  heating  ? 

2.  How  much  of  the  oxide  must  be  used  in  order  to 
obtain  30  grammes  of  oxygen  ? 

IVInOa  +  HO,  ^S  C>3  =  MnO,  S  O3  4-  HO  +  O. 

1.  How  much  oxygen  can  be  obtained  from  one  kilogr. 
of  hyperoxide  of  manganese  by  the  last  process  ?  and  how 
much  sulphuric  acid  wiU  be  required  to  decompose  it  ? 

2.  "What  per  cent  of  the  whole  amount  of  oxygen  in 
the  mineral  is  obtained  by  the  two  processes  ?  and  by  how 
much  does  the  second  exceed  the  first  ? 


NON-METALLIC    ELEMENTS,    OR    METALLOIDS.  49 

Hydrogen  (H). 

81.  Na  +  ^g  =  Na  0,  HO^Aq-\-  H. 

1.  How  much  hydrogen  will  be  set  free  by  5.25 
grammes  of  sodium?  How  many  cubic  centimetres  will 
be  set  free? 

82.  3  Fe  +  4  M  ©  ==  Fe  O,  Fcj  O3  +  4  H. 

1.  How  much  hydrogen  would  be  obtained  in  the  last 
experiment  by  the  decomposition  of  39  grammes  of 
water  ? 

2.  By  how  much  would  the  weight  of  the  tube  increase 
in  last  example  ? 

83.  Zn  +  HO,  S  0^^  Aq  =  Zn  0,  S  0^-^  Aq  -\-ll. 

1.  How  much  sulphuric  acid  and  how  much  zinc  must 
be  used  in  order  to  make  2  grammes  of  hydrogen  ?  How 
much  in  order  to  make  one  litre  ? 

2.  How  much  hydrogen  can  be  obtained  with  one  kilo- 
gramme of  zinc  ? 

3.  How  much  with  one  kilogramme  of  sulphuric  acid  ? 

Decomposed  by  Galvanism. 

55.  iro  =  H  +  ©. 

1.  How  many  cubic  centimetres  of  hydrogen  would  be 
obtained  by  the  decomposition  of  one  cubic  centimetre 
(one  gramme)  of  water  ?  How  many  cubic  centimetres 
of  oxygen  ?  How  many  of  mixed  gases  ? 
85.  A  solid  immersed  in  a  liquid  or  a  gas  is  buoyed  up 
by  a  force  equal  to  the  weight  of  the  liquid  or  gas  which 
it  displaces.  The  excess  of  the  buoyancy  over  its  own 
weight  is  called  its  ascensional  force. 

1.  What  would  be   the   ascensional  force  of  a  small 
balloon  filled  with  one  htre  of  hydrogen  gas,  when  the 
balloon  itself  weighs  five  centigrammes  ? 
5 


50  NON-METALLIC   ELEMENTS,    OR   METALLOIDS. 

2.  What  would  be  the  ascensional  force  of  a  spherical 
balloon  seven  metres  in  diameter,  two  thirds  filled  with 
hydrogen,  when  the  balloon  and  attachments  weigh  twenty 
kilogrammes  ? 

87.  n-i-o  =  iro. 

1.  How  much  water  would  be  formed  by  burning  one 
thousand  litres  of  hydrogen?  and  how  much  oxygen 
would  be  consumed  in  the  process  ? 

2.  How  much  vapor  of  water  would  be  formed  in  Ex- 
ample 1  ? 

93.  Problems  on  the  Barometer. 

1.  When  the  surface  of  the  column  of  mercury  in  a 
barometer  stands  at  76  centimetres  above  the  mercury  in 
the  basin,  with  what  weight  is  the  atmosphere  pressing  on 
every  square  centimetre  of  surface  ?  Sp.  Gr.  of  mercury 
=  13.596. 

2.  To  what  difference  of  pressure  does  a  difference  of 
one  centimetre  in  the  barometric  column  correspond  ? 

3.  When  the  water  barometer  stands  at  ten  metres, 
what  is  the  pressure  of  the  air  if  the  temperature  is  4°  ? 

4.  How  high  would  an  alcohol  barometer,  and  how 
high  a  sulphuric-acid  barometer,  stand  under  the  same 
circumstances,  disregarding  in  each  case  the  tension  of 
the  vapor  ?  Sp.  Gr.  of  alcohol  =  0.8095  ;  Sp.  Gr.  of 
sulphuric  acid  =  1.85. 

94.  Prohlems  on  the  Compression  and  Expansion  of  Gases. 

Mariotte's  Law.  —  It  is  an  estabhshed  principle  of  sci- 
ence that  I'he  volume  of  a  given  weight  of  gas  is  inversely 
as  the  pressure  to  which  it  is  exposed  ;  that  is,  the  greater 
the  pressure,  the  smaller  is  the  volume ;  and  the  less  the 


NON-METALLIC    ELEMENTS,    OR   METALLOIDS.  51 

pressure,  the  larger  is  the  volume.  This  may  be  illustrated 
by  an  India-rubber  bag  holding  one  litre  of  air,  or  of  any 
other  gas.  This  is  exposed  to  a  pressure,  under  the  ordi- 
nary conditions  of  the  atmosphere,  of  a  little  over  one 
kilogramme  on  every  square  centimetre  of  surface.  If 
this  pressure  is  doubled,  the  volume  of  the  bag  will  be 
reduced  to  one  half;  if  trebled,  to  one  third,  &c.  On 
the  other  hand,  if  the  pressure  is  reduced  to  one  half, 
the  volume  will  double ;  if  to  one  third,  the  volume  will 
treble,  &c.*  The  principle  is  expressed  in  mathematical 
language  by  the  proportion 

-  H'  :  H=:  V  :  V;  (1.) 

where  H  and  H'  are  the  heights  of  the  barometer  which 
measure  the  pressure  to  which  the  gas  is  exposed  under 
the  two  conditions  of  volume  V  and  V. 

Since  the  density  of  a  given  weight  of  gas  is  inversely 
as  the  volume,  or  D' :  D  =  V  :  V,  it  follows  that 

H' :  H  =  D' :  D,  (2.) 

or  the  density  of  a  gas  is  proportional  to  the  pressure  to 
which  it  is  exposed.  Moreover,  since  the  weight  of  a 
given  volume  of  gas  is  proportional  to  its  density,  and  its 
density,  as  just  proved,  proportional  to  the  pressure,  it 
follows  that  The  weight  of  a  given  volume  of  gas  is  directly 
as  the  pressure  to  which  it  is  exposed,  or 

H'  :  H  =  W  :  W.  (3.) 

These  three  proportions  are  very  important,  and  will  be 
constantly  referred  to  in  the  following  pages.  The  student 
must  be  careful  to  notice  that  in  (1)  the  weight  of  gas  is 
supposed  to  be  constant  and  the  volume  to  vary,  and  in 
(g)  the  volume  is  supposed  to  be  constant  and  the  weight 
to  vary. 

*  We  suppose  the  bag  to  have  no  elasticity. 


52  NON-METALLIC    ELEMENTS,    OR   METALLOIDS. 

The  variations  in  the  pressure  of  the  atmosphere,  amount- 
ing at  times  to  one  tenth  of  the  whole,  necessarily  cause 
equally  great  changes  in  the  volume  of  gases  which  are 
the  objects  of  chemical  experiment.  Hence,  in  order  to 
compare  together  different  volumes  of  gas,  it  is  essential 
that  they  should  have  been  measured  when  exposed  to 
the  same  pressure.  A  standard  pressure  has  therefore 
been  agreed  upon,  that  measured  by  76  centimetres  of 
3iiercury,  to  which  the  volume  of  gases  measured  under 
any  other  pressure  should  be  reduced.  Hence  a  number 
of  problems  like  the  following :  — 

1.  A  volume  of  hydrogen  gas  was  measured  and  found 
to  be  equal  to  250  c.  c.  The  height  of  the  barometer, 
observed  at  the  same  time,  was  74.2  centim.  What 
would  have  been  the  volume  if  observed  when  the  barom- 
eter stood  at  76  centim.  ? 

Solution.  —  Proportion  (1)  gives,  by  substituting  the  data  of  the  prob- 
lem, 74.2  :  76  =  250  :  V  =  Ans. 

2.  A  volume  of  nitrogen  gas  measured  756  c.  c.  when 
the  barometer  stood  at  77.4  centim.  What  would  it  have 
measured  if  the  barometer  had  stood  at  76  centim.  ? 

3.  A  volume  of  air  standing  in  a  bell-glass  over  a  mer- 
cury pneumatic  trough  measured  568  c.  c.  The  barome- 
ter at  the  time  stood  at  75.4  centim.,  and  the  surface  of  the 
mercury  in  the  bell  was  found,  by  measurement,  to  be  6.5 
centim.  above  the  surface  of  the  mercury  in  the  trough. 
What  would  have  been  the  volume  had  the  air  been  ex- 
posed to  the  pressure  of  76  centim.  ? 

Solution.  —  It  can  easily  be  seen,  that  the  pressure  of  the  air  on  the 
surface  of  the  mercury  in  the  pneumatic  trough,  measured  by  the 
height  of  the  barometer  at  the  time  (75.4  centim.),  was  balanced  first  by 
the  column  of  mercury  in  the  bell,  and  secondly  by  the  tension  *  of  the 
confined  air.    Hence,  the  pressure  *  to  which  the  air  was  exposed  was 

*  The  tension  of  a  gas  is  the  force  with  which  it  tends  to  expand,  and,  when 
the  gas  is  at  rest,  must  evidently  be  exactly  equal  to  the  pressure  to  which  it 
is  exposed. 


NON-METALLIC   ELEMENTS,    OR   METALLOIDS.  53 

equal  to  the  height  of  the  barometer  less  the  height  of  the  mercury 
in  the  bell,  or  75.4  —  6.5  =  68.9  centim.  We  have  then  the  proportion 
68.9  :  76  =  568  :  V  =  Ans. 

4.  A  volume  of  air  standing  in  a  tall  bell-glass  over  a 
mercury  pneumatic  trough  measured  78  c.  c.  The  barom- 
eter at  the  time  stood  at  74.6  centim.,  and  the  mercury  in 
the  bell  at  57.4  centim.  above  the  mercury  in  the  trough. 
"What  would  have  been  the  volume  had  the  pressure  been 
76  centim.  ? 

5.  What  would  be  the  answers  to  the  last  two  problems, 
had  the  pneumatic  trough  been  filled  with  water  instead 
of  mercury  ? 

97.  Problems  on  Expansion  of  Gases  hy  Heat. 

1.  What  will  be  the  volume  of  250  cubic  centimetres 
of  air  at  0°  when  heated  to  300°  ? 

Solution.  —  It  has  been  found  by  Eegnault  and  others,  that  the  per- 
manent gases  expand  so  nearly  equally  for  the  same  increase  of  tempera- 
ture, that  the  differences  may  be  entirely  disregarded  except  in  the  most 
refined  investigations;  and  it  has  also  been  found,  that  their  rate  of 
expansion  does  not  materially  vary  from  the  lowest  to  the  highest  tem- 
peratures at  which  experiments  have  been  made.  The  coefficient  of 
expansion  for  air,  as  determined  by  Eegnault,  is  equal  to  0.00366,  and 
we  can  therefore  calculate  the  volume,  V,  of  a  gas  at  any  temperature, 
from  its  volume,  V,  at  0°,  by  means  of  the  equation,  already  explained, 

V'  =  V(l  +  iX  0.00366);  (1.) 

or  if  we  know  the  volume  Y'  for  a  tempei-atiire  t,  we  can  calculate  the 
volume  V"  for  another  temperature  i\  by  means  of  the  equation 

V"  =  V' (!  +  («'  —  <)  0.00366).  (2.) 

By  transposing,  we  can  obtain  from  equation  (1) 

l  +  tX  0.00366'  ^     ' 

As  the  volume  of  a  gas  varies  very  considerably  with  the  temperature, 
it  is  important,  in  comparing  together  different  measurements,  that  we 
should  adopt  a  standard  temperature,  as  we  have  adopted  a  standard 
pressure.  The  temperature  which  has  been  agreed  upon  is  0° ;  but  as  it 
would  be  inconvenient,  and  often  impossible,  to  make  our  measurements 
at  this  temperature,  it  becomes  necessary  to  calculate,  by  means  of  equa- 
5* 


54  NON-METALLIC    ELEMENTS,    OR   METALLOIDS. 


tion  (3),  from  a  volume  Y'  measured  at  t°,  what  would  be  the  volume  at 
0°.  This  is  called  technically  reducing  the  volume  to  0°.  There  can  be 
obtained  also  from  equations  (1)  and  (2)  the  equations 

_      V''  — V  ,  _  V"  —  V 

'  ~  V  X  0.00366 '        ^^-^  ^^^      *  ~  *  "^  V  X  0.00366 '  ^^'^ 

by  means  of  which  we  can  calculate  the  change  of  temperature  when 
we  know  the  change  of  volume.  Kepresenting  the  coefficient  of  expan- 
sion in  the  above  formulte  by  Jc,  we  can  obtain,  by  transposing  and  re- 
ducing the  equation, 

V'-V  ,  ,      ,       V"  — V 

''  =  -lir'        (60         and     ^  =  v' (£/-«)•  CO 

From  these  we  can  calculate  the  coefficient  of  expansion  when  we  know 
the  volume  of  a  gas  at  two  diflfereat  temperatures. 

2.  A  volume  of  gas  measured  560  c.  c.  at  15°.  What 
would  it  measure  at  95°  ? 

3.  A  glass  globe  holding  450  c.  c.  of  air  at  0°  was 
heated  to  300°.  At  this  temperature  the  neck  was  her- 
metically sealed,  and  the  globe  cooled  again  to  0°.  The 
neck  was  then  opened  under  mercury,  and  the  air  remain- 
ing in  the  globe  passed  up  into  a  graduated  jar,  and  meas- 
ured.    How  much  was  it  found  to  measure  ? 

Solution.  —  By  substituting  the  values   for  V  and  t  in  the  equation 

V  =  V  (1  +  <  X  0.00003),  we  obtain  the  increased  capacity  of  the  globe, 
and  of  course  the  number  of  cubic  centimetres  of  expanded  air  which  it 
contains  at  300^.    It  is  then  only  necessary  to  substitute  this  value  for 

V  and  300  for  t,  in  equation  (3),  in  order  to  find  what  will  be  the  volume 
of  this  expanded  air  when  cooled  again  to  0°. 

4.  "What  is  the  weight  of  air  contained  in  an  open  glass 
globe  of  250  c.  c.  capacity,  at  the  temperature  of  20°,  and 
when  the  barometer  stands  at  74  centimetres  ? 

Solution.  —  In  order  to  make  the  solution  general,  we  will  represent  the 
capacity  of  the  globe,  the  temperature,  and  the  height  of  the  barometer, 
by  V,  t,  and  H,  respectively.  One  cubic  centimetre  of  air  at  0°,  and 
when  the  barometer  stands  at  76  centimetres,  weighs  0.00129  grammes. 
To  find  what  one  cubic  centimetre  would  weigh  when  the  barometer 
stands  at  H  centimetres,  we  make  use  of  proportion  (3),  page  51: 

H  :  H'  =  W  :  W;      or      76  :  H  =  0.00129  :  W; 

whence  W  =  0.00129  .  ^,  the  weight  of  one  cubic  centimetre  at  0^,  and 
under  a  pressure  of  H  centimetres.    To  find  what  one  cubic  centimetre 


NON-METALLIC    ELEMENTS,    OR   METALLOIDS. 


55 


would  weigh  at  P,  it  must  be  remembered  that  one  cubic  centimetre  at  0= 
becomes  (1  ■+•  t  0.00366)  c.  c.  at  t^;  therefore,  at  P  and  at  H  centimetres 
of  the  barometer,  (1  +  i  0.00366)  c.  c.  weigh  0.00129  .  ^  grammes.  By 
equating  these  two  terms  we  obtain  (1  + 1  0.00366)  =  0.00129  .  ^,  whence 
1  =  0.00129  •  ]-^ x7o"oo366  '  76'  *^®  weight  of  one  cubic  centimetre  at  P  and 
under  a  pressure  of  H  centimetres.  The  weight  of  V  cubic  centimetres, 
w,  is  evidently 


zi)  =  0.00129  V  . 


H 


1  +  (  0.00366     76  * 


(8.) 


5.  Wliat  is  the  weight  of  air  contained  in  an  open  glass 
globe  of  560  cubic  centimetres'  capacity  at  0°,  at  the 
temperature  of  300°,  and  under  a  pressure  of  77  centi- 
metres ? 

Solution.  —  In  the  soli;tion  of  the  last  example  we  neglected  the  change 
of  capacity  of  the  glass  globe  due  to  the  change  of  temperature.  This 
causes  no  sensible  error  when  the  change  of  temperature  is  small,  but 
when,  as  in  the  present  problem,  the  change  of  temperature  is  quite 
large,  the  change  of  capacity  of  the  globe  must  be  considered.  If  the 
capacity  is  V  c.  c.  at  0^,  it  becomes  atPY{l  +  t  0.00003).  Introducing 
this  value  for  V  into  the  equations  of  the  last  section,  we  obtain 


w  =  0.00129  V  (1  +  <  0.00003) 


1  +  t  0.0U366      76  ■ 


(9.) 


99.  Problems  on  Specific  Gravity  of  Vapors. 

General  Solution.  —  The  specific  gravity  of  a  vapor  is  its  weight  com- 
pared with  the  weight  of  the  same  volume  of  air  under  the  same  con- 
ditions of  temperature  and  pres- 


sure. To  find,  then,  the  specific 
gravity  of  a  vapor,  we  must  as- 
certain the  weight  of  a  known 
volume,  V,  at  a  known  tempera- 
ture, t,  and  under  a  known  pres- 
sure, H,  and  divide  this  by  the 
weight  of  the  same  volume  of  air 
at  the  same  temperature,  and 
under  the  same  pressure.  The 
method  may  best  be  explained  by 
an  example.  Suppose,  then,  that 
we  wish  to  ascertain  the  specific 
gravity  of  alcohol  vapor.  We  take 
a  light  glass  globe  having  a  ca- 
pacity of  from  400  to  500  c.  c,  and 


Fig.  6. 


56  NON-METALLIC    ELEMENTS,    OR  METALLOIDS. 

draw  the  neck  out  in  the  flame  of  a  blast  lamp,  so  as  to  leave  only  a  fine 
opening,  as  shown  in  the  figure  at  a.  The  first  step  is  now  to  ascertain 
the  weight  of  the  glass  globe  when  completely  exhausted  of  air.  As 
this  cannot  readily  be  done  directly,  we  weigh  the  globe  full  of  air,  and 
then  subtract  the  weight  of  the  air,  ascertained  by  calculation  from  the 
capacity  of  the  globe,  and  from  the  temperature  and  pressure  of  the  air, 
by  means  of  equation  (8).  Call  the  weight  of  the  globe  and  air  W,  and 
the  weight  of  the  air  w,  then  W  —  w  is  the  weight  of  the  globe  exhausted 
of  air.  The  second  step  is  to  ascertain  the  weight  of  the  globe 
filled  with  alcohol  vapor  at  a  known  temperature,  and  under  a  known 
pressure.  For  this  purpose  we  introduce  into  the  globe  a  few  grammes 
of  pure  alcohol,  and  mount  it  on  the  support  represented  in  Fig.  6.  By 
loosening  the  screw,  r,  we  next  sink  the  balloon  beneath  the  oil  contained 
in  the  iron  vessel,  V,  and  secure  it  in  this  position.  We  noAV  slowly  raise 
the  temperature  of  the  oil  to  between  300°  and  400°,  which  we  observe  by 
means  of  the  thermometer,  B.  The  alcohol  changes  to  vapor  and  drives 
out  the  air,  which,  with  the  excess  of  vapor,  escapes  at  a.  When  the 
bath  has  attained  the  requisite  temperatui'e,  we  close  the  opening  a,  by 
suddenly  melting  the  end  of  the  tube  at  a  with  a  mouth  blow-pipe,  and 
as  nearly  as  possible  at  the  same  moment  observe  the  temperature  of  the 
bath  and  the  height  of  the  barometer.  We  have  now  the  globe  filled  with 
alcohol  vapor  at  a  known  temperature,  and  under  a  known  pressure. 
Since  it  is  hermetically  sealed,  its  weight  cannot  change,  and  we 
can  therefore  allow  it  to  cool,  clean  it,  and  weigh  it  at  our  leisure. 
This  will  give  us  the  weight  of  the  globe  filled  with  alcohol  vapor 
at  a  known  temperature,  t',  and  under  a  known  pressure,  H'.  Call 
this  weight  W'.  The  weight  of  the  vapor  is  W'  —  W  +  w.  The 
third  step  is  to  ascertain  the  weight  of  the  same  volume  of  air  at  the 
same  temperature  and  under  the  same  pressure.  This  can  easily  be 
found  by  calculation  from  equation  (9).  The  last  step  is  to  find  the 
capacity  of  the  globe,  which,  although  we  have  supposed  it  known,  is 
not  actually  ascertained  experimentally  until  the  end  of  the  process. 
For  this  purpose  we  break  off  the  tip  of  the  tube  (a),  under  mercury, 
which,  if  the  experiment  has  been  carefully  conducted,  rushes  in  and 
fills  the  globe  completely.  We  then  empty  this  mercury  into  a  carefully 
graduated  glass  cylinder,  and  read  oS"  the  volume.  We  find  then  the 
specific  gravity  by  dividing  the  weight  of  the  vapor  by  the  weight  of 
the  air.    The  formula  for  the  calculation  are  then 

Weight  of  the  globe  and  air,  W. 

"        "        air,  ,„  =  0.00129  V.pp-^^.  5. 

"        "        globe  exhausted  of  air,      W  —  w. 

"        "  "     filled  with  vapor  '\ 

at  a  temperatm-e  t'  and  under  a  >  W. 

pressure  H',  ) 


NON-METALLIC    ELEMENTS,    OR   METALLOIDS. 


57 


Weight  of  the  vapor,  W  —  W  +  to. 

"         "         air   a.t   t'    )  1  H 

ail   HI  I    (  =,0.00129  V(l  +  «' 0.00003)-   i  + ,.  o.uosss  '  76 ' 
jsure  H',  )  ^ 

W  —  W  +  w 

Sp.  Gr.  = . 


and  under  a  pressure 


0.00129  V  (l  +  C  0.00003). 


1  +  t  0.00366     75 


1.  Ascertain  the  Sp.  Gr.  of  alcohol  vapor  from  the  fol- 
lowing data :  — 


"Weight  of  glass  globe,           W 

50.8039  grammes. 

Height  of  barometer,             H 

74.754    centim. 

Temperature,                           t 

18° 

Weight  of  globe  and  vapor,  W 

50.8245  grammes. 

Height  of  barometer,             H' 

74.764    centim. 

Temperature,                          i' 

167° 

Volume,                                 V 

351.5      cubic  centim. 

Ans.  1.5795. 

2.  Ascertain  the  Sp.  Gr.  of 

camphor  vapor  from  the 

following  data :  — 

Weight  of  glass  globe,           W 

50.1342  grammes. 

Height  of  barometer,             H 

74.2        centim. 

Temperature,                           t 

13°.5 

Weight  of  globe  and  vapor,  W 

50.8422  grammes. 

Height  of  barometer,             H' 

74.2         centim. 

Temperature,                           t 

244° 

Volume,                                 V 

295        cubic  centim. 

Ans.  5.298. 

Carbon  (C). 

109.  C  +  20  =  C02. 

2  Hg  0  4-  C  =  2  ^r/  -}-  C  ©2- 

1.  HoAV  many  grammes  and  how  many  cubic  centi- 
metres of  carbonic  acid  gas  are  formed  by  burning  10 
grammes  of  charcoal  ? 


58     ■         NON-METALLIC   ELEMENTS,   OR   METALLOIDS. 

2.  Ho"W  many  grammes  and  hoTV  many  cubic  centi- 
metres of  oxygen  are  consumed  in  the  process  ? 

3.  Assuming  that  the  volume  of  carbonic  acid  gas  gen- 
erated during  combustion  is  exactly  equal  to  the  volume  of 
oxygen  gas  consumed,  what  is  the  Sp.  Gr.  of  carbonic 
acid  gas  ? 

4.  How  much  oxide  of  mercury  is  required  to  bum  up 
5.672  grammes  of  charcoal  ? 

110.  C  +  0  =  €0.  C  +  C02=2CO. 

1.  How  many  grammes  and  how  many  cubic  centime- 
tres of  oxide  of  carbon  gas  are  formed  by  burning  10 
grammes  of  charcoal  ? 

2.  How  many  grammes  and  how  many  cubic  centime- 
tres of  oxygen  are  consumed  in  the  process  ? 

8.  Ten  cubic  centimetres  of  oxygen  yield  how  many 
cubic  centimetres  of  carbonic  acid  gas,  and  how  many  of 
oxide  of  carbon  gas  ?  What  expansion  does  oxygen  un- 
dergo in  combining  with  carbon  to  form  oxide  of  carbon  ? 

Spermaceti. 

115.  C64  H64  O4  +  188  O  =  64  C  O2  +  64  HO. 

1.  How  many  grammes  of  carbonic  acid,  and  how  many 
grammes  of  water,  are  formed  by  burning  10  grammes  of 
spermaceti  ? 

2.  The  carbonic  acid  and  water  given  off  by  a  burning 
spermaceti  candle  were  carefully  collected  and  weighed. 
The  water  weighed  0.564  grammes,  the  carbonic  acid 
weighed  1.3786.     How  much  of  the  candle  was  burned? 


NON-METALLIC   ELEMENTS,    OR   METALLOIDS.  59 

SECOND   GROUP   OF  METALLOIDS. 

Sulphur  (S). 

132.  Fe  S  +  HO,  80^-\-Aq^Fe0,S0^^Aq•\-  H  S. 

1.  How  much  sulphide  of  hydrogen  can  be  made  from 
15  grammes  of  sulphide  of  iron  ?  How  many  cubic  centi- 
metres ? 

2.  How  much  sulphide  of  iron,  and  how  much  sulphuric 
acid,  is  required  to  generate  sufficient  gas  to  saturate  one 
litre  of  water  ? 

HS-^Aq-\-i^=^-\-Aq. 
HS  +  30=  eO  +  SOij. 

Black. 

133.  a.  Pb  +  iT^-f  ^^  =  Pb  S  +  ^^  +  H. 

Yellow.  Black. 

l.YhO  ^  H  S  -\-  Aq  =  Y'o^  -\-  Aq. 

Black. 

c.PlO,    {_G^H;\  O3  4-   H8  -\-  Aq    =    PbS    -f 
HO,lG^H,-\0,JrM- 

Black. 

d. Fe  0,S0^-\-  GaO,HO-\-  HS  -{-Aq^YQ^  + 
Ga  0,S0^-\-  Aq. 

Black. 

Gu  0,  [O;  H,-]  0,    -{.    ITS  +   Aq  =    Cu  S  + 
irO,[G,ff,-]  0,  +  Aq. 

Orang'e# 

Sb  Gl3-}-3JIS-\-Aq  =  ^hS3-{-3irGl-{-  Aq. 

Yellow. 

As  Gls  -i-SIIS-^  Aq  =  As  S3 -\-  dHGl-\-Aq. 

White. 

ZnGl-{-GaSA  Aq  =ZnS-\-  Ga  Gl-\- Aq. 


60  NON-METALLIC   ELEMENTS,    OR  METALLOIDS. 

Phosphorus  (P). 

140.       P  +  3  O  =  P  O3  (by  slow  combustion). 
P-f-5O==P05(by  rapid  combustion). 

144.  Burnt  bones  consist  chiefly  of  3  Ca  O,  cP  O5. 

3  Ca  O,  cP  O5  +  2  {HO,  S0,)-\-aq  =  2  (CaO,  S  O3) 
+  [2  HO,  Ca  01  eP  O5  +  aq. 

Heated  to  a  red  heat. 

\^H0,  Ca  0],  ^PO,  ^  Aq  -\-xC  =  CaO,  PO5 

+  a;C  +  Aq  +  2CO  +  2H. 

Heated  intensely. 

3  (CaO,  ,P Os)  +  a;  C  =  3  CaO,  aPOs  +  a:  C  +  2  P 
+  10  C  O.  ^ 

1.  How  much  phosphorus  can  be  manufactured  from 
20  kilogrammes  of  burnt  bones,  of  which  four  fifths  are 
phosphate  of  lime  ? 

145.  4P  +  3(CaO,  HO)  =  3(CaO,  P0)  +  PH3. 

Besides  the  above  reaction,  there  take  place  simulta- 
neously the  two  following  reactions,  in  the  experiment 
described  in  the  text-book. 

3  P  4-  2  (Ca  0,  H  0)  =  2  (Ca  0,  P  0)  +  P  H^. 

P  -f  Ca  O,  H  0  =  Ca  0,  P  0  +  M. 


THIRD    GROUP   OF   METALLOIDS. 

Chlorine  (CI). 

150.  Mn  0.,-^2HGl-\-aq  =  Mn  CI -{- aq -{-  CI. 

1.  HoAv  much  chlorine  gas  can  be  obtained  from  2.467 


NON-METALLIC   ELEMENTS,    OR   METALLOIDS.  61 

grammes  of  clilorohydric  acid  gas  ?     How  many  cubic 
centimetres  ? 

2.  How  much  chlorine  can  be  obtained  from  an  unde- 
termined amount  of  muriatic  acid  by  means  of  4.567 
grammes  of  hyperoxide  of  manganese  ?  How  many  cubic 
centimetres  ? 

3.  The  hyperoxide  of  manganese  of  commerce  is  more 
or  less  adulterated.  What  per  cent  of  Mn  O2  does  an 
article  contain,  of  which  10  grammes,  when  heated  with 
strong  muriatic  acid,  evolve  4.0135  grammes  of  chlorine  ? 

4.  How  much  chlorine  can  be  obtained  from  25  cubic 
centimetres  of  muriatic  acid*  of  Sp.  Gr.  =  1.16  ?  How 
many  cubic  centimetres  ? 

5.  In  order  to  prepare  one  litre  of  chlorine  gas  how  much 
hyperoxide  of  manganese,  and  how  much  muriatic  acid, 
must  be  used  ?  Calculate  the  amounts  for  pure  Mn  O2  and 
H  CI  gas,  and  also  when  the  oxide  used  contains  only 
70  per  cent  of  pure  MnOa,  and  when  the  liquid  acid 
used  has  a  Sp.  Gr.  =  1.15. 

151.    Mn  O2  +  2  Na  CI  +  2  (^  0,  aS'  O3)  -{- aq  =  Mn  a  + 
2  (JVa  0,  S  Os) -{- ag -{- Ch 

"We  might  use  one  half  as  much  common  salt,  but  then 
we  should  find  sulphate  of  manganese  instead  of  chloride 
of  manganese  in  solution.     Thus, 

Mn  O2  +  Na  CI  +  2  {HO,  S  0^)  -{-aq=-.Mn  0,  SO, 
Jf-J^aO,SOs-\-aq^  CI. 

1.  How  much  chlorine  gas  can  be  obtained  by  the  last 
process  from  34  kilogrammes  of  salt  ? 

*  See  Table  VI.,  which  gives  the  per  cent  of  H  CI  in  the  fluid  acid  of  dif- 
ferent specific  gravities. 

6 


62  NON-METALLIC    ELEMENTS,    OR   METALLOIDS. 

2.  How  many  cubic  centimetres  of  chlorine  can  be  ob- 
tained from  one  cubic  centimetre  of  rock  salt  ?  Sp.  Gr. 
of  salt  =  2.15. 

152.  /.  2  {Fe  0,  S  0^)  -{- H 0,  S  0^  -\-  Gl -\- Aq  =  Fe^  0^, 
3  S  Os  +  II  CI -\- Aq. 

6  (Fe  0,8  0^)  -\-  3  CI  -\-Aq  =  2  {Fe.^  O3,  3  S  O3) 
_|_  Fe^  Ck  +  Aq. 

g.  An -^  3  CI -\- Aq  =^  An  CI,  +  Aq. 


ACIDS 


FIRST    GROUP  :    OXYGEN   ACIDS. 

Nitrogen  and  Oxygen. 

Nitric  Acid  {HO,  NO,). 
159.  KO,  N  Os  +  2  {HO,  8  0^)  =  KO,  SO,.  HO,  S  O3 

NaO,  NO5  +  2  {HO,  SOs)  =  NaO,  SO,.  HO,  S  0, 

+  H0,J¥05. 

1.  How  much  nitric  acid  can  be  made  from  250  kilo- 
grammes of  potash  nitre,  and  how  much  sulphuric  acid 
must  be  used  in  the  process  ? 

2.  How  much  more  nitric  acid  will  the  same  weight  of 
soda  nitre  yield  ? 

3.  How  much  nitric  acid,  containing  40  per  cent,  of 
NO5,  can  be  made  from  1700  kilogrammes  of  potash 
nitre  ? 

4.  How  much  soda  nitre,  and  how  much  sulphuric  acid, 
and  how  much  water,  must  be  used  to  make  450  kilo- 
grammes of  nitric  acid,  which  shall  contain  60  per  cent. 
of  pure  acid  ? 


64  ACIDS. 

Ammonia, 

160.  c.  INH,-]  0,  HO  +  HO,  NO,  -^Aq^  iNH,-\  0,  N  0, 

+  Aq. 

d.VhO-{-HO,NO,-\-Aq  =  PhO,NOs-\-Aq. 

1.  How  much  nitric  acid,  of  Sp.  Gr.  1.14,  is  required 
to  dissolve  20  kilogrammes  of  oxide  of  lead  ? 

2.  How  much  nitric  acid,  of  Sp.  Gr.  1.14,  and  how 
much  oxide  of  lead,  must  be  used  to  make  10  kilogrammes 
of  nitrate  of  lead  ? 

c.  3  Pb  +  4  {HO,  N O5) -\- aq  =^  Z  (Pb  0,  NO5)  +  aq 

+  1V02. 

3  Cu  +  4  {HO,  NO,)^aq=  3  {Cu  0,  NO,)  +  aq 

1.  How  much  nitric  acid,  of  Sp.  Gr.  1.22,  is  required 
to  dissolve  450  grammes  of  lead  ?  How  much  nitrate  of 
lead  would  be  formed  ? 

2.  How  much  nitric  acid,  of  Sp.  Gr.  1.362,  is  required 
to  dissolve  450  grammes  of  copper  ? 

/.  3  P  +  5  (^  (9,  NO,)  -{-Aq=3{3HO,,P  0,)  +  Aq 
+  5  ]^  O2. 
S  +  HO,  NO,  -\-Aq  =  HO,SOs  +  Aq  +  JVO^. 

Nitric  Oxide  (X  O). 

6FeCl-\-KO,N05-{-  4.H01  -{- Aq  =  S  Fe^  CI3  + 

KCl  +  aq-\-NO!i. 

Colorless.  Red. 

1.  How  much  IV  O2  can  be  obtained  by  dissolving  10 
grammes  of  copper  in  nitric  acid  ?  How  many  cubic 
centimetres  ? 


ACIDS.  65 

2.  How  much  IV  O2  can  be  obtained  from  10  grammes 
of  iron  by  the  last  reaction  but  one  ? 

3.  "What  volume  of  oxygen  must  be  mixed  with  one 
litre  of  TV  O2  in  order  to  change  it  into  ]V  O4  ? 

Nitrous  Oxide  (IV  O). 

When  heated. 

163.  [NHi]0,  N05=4HO  +  2]VO. 

Sp.Gr.  of  1.14. 

4.Vh^b{H0,  N0,)+Aq  =  4:  (Pb  0,  NOs)  +  Aq 

+  JVO. 

Mi  0,  S  0^  -\-  Aq  -{-  -X  0^  =  Na  0,  S  0^  -{-  Aq  -\- 
IV©. 

1.  Ten  grammes  of  nitrate  of  ammonia  yield  how  many 
grammes  and  how  many  cubic  centimetres  of  protoxide  of 
nitrogen  ? 

2.  How  much  nitrate  of  ammonia  must  be  used  in  order 
to  make  one  litre  of  the  gas  ? 

3.  One  litre  of  TV  O2  yields  how  many  cubic  centime- 
tres of  IV  O  by  the  third  reaction  of  this  section  ? 

4.  One  litre  of  IVO  gives,  when  decomposed,  what 
volume  of  nitrogen  ? 

Carbon  and  Oxygen. 
Carbonic  Acid  (C  02)' 

164.  CaO,  CO2  4-  HO,  NO,  -\-  Aq  =  Ca  0,  NO^  +  Aq 

+  CO2. 

Ca 0,  C O2  +  ^0,  S 0^-\-  Aq=  C^O,  S O3  +  Aq 

+  C  O2. 

6* 


•66  ACIDS. 

Ca  0,  NOs  +  HO,  SO3  +  aq   =    C?.0,   ^0^-{- 
HO,NOs-\-aq. 

1.  How  much  sulphuric  acid  and  how  much  nitric 
acid*  must  be  used  to  drive  out  all  the  carbonic  acid  from 
25.462  grammes  of  chalk  ?  How  many  grammes  and  how 
many  cubic  centimetres  of  gas  would  be  obtained  ? 

2.  The  specific  gravity  of  Carrara  marble  is  2.716. 
How  many  cubic  centimetres  of  carbonic  acid  gas  does 
one  cubic  centimetre  of  the  marble  contain  in  a  condensed 
state  ? 

167.  1.  Animals  remove  oxygen  from  the  air,  and  return  the 
whole  as  carbonic  acid.  Plants  remove  carbonic  acid, 
and,  having  decomposed  it,  return  the  oxygen  it  contained. 
How  does  the  volume  of  the  oxygen  in  either  case  com- 
pare with  that  of  the  carbonic  acid  ? 

Sulphur  and  Oxygen. 
Sulphuric  Acid  (S  O3). 

169.  S02  4-Ot=  SO3. 

1.  How  much  anhydrous  sulphuric  acid  is  formed  by 
the  oxidation  of  10  grammes  of  sulphurous  acid?  and 
how  much  oxygen  is  required  in  the  process  ? 

2.  How  much  anhydrous  sulphuric  acid  is  formed  by 
the  oxidation  of  one  litre  of  sulphurous  acid  gas  ?  and 
how  many  cubic  centimetres  of  oxygen  must  be  mixed 
with  it  in  the  experiment  ? 

a.  2  S  O3  +  H  O  =  iT  0,  2  ^  ^3  (the  Nordhausen  Acid). 

*  When  the  strength  of  the  nitric  acid  is  not  stated,  monohydrated  acid 
(H  0,  N  O5)  is  always  intended. 

t  The  two  gases  are  mixed  together  and  led  over  heated  platinum  sponge  in 
a  glass  tube. 


ACIDS.  67 

S  O3  +  H  O  =  ^  0,  aS'  6>3  (the  common  Acid). 

170.  Fe  O,  S  O3 .  6  H  0  is  the  symbol  of  ciystallized  green 

vitriol. 

When  healed. 

2(FeO,  SO3  .  6HO)  =  Fe203,  S  O3  +  12IIO 

+  S©2. 
By  further  heating", 

Fe2  03,  S03=Fe2  03  +  §©3. 

By  conducting  the  anhydrous  acid  fumes  into  HO,  S  O3 
we  get  IIO,2S03. 

1.  How  much  anhydrous  acid,  and  how  much  Nord- 
hausen,  can  be  made  by  the  above  process  from  20  kilo- 
grammes of  green  vitriol  ?  If  the  Nordhausen  acid  has 
the  specific  gravity  of  1.9,  how  many  litres  can  be  obtained 
from  20  kilogrammes  of  green  vitriol  ? 

171.  S02  +  M.0,N0s  +  Afl  =  HO,  SOs  +  Aq-^NOi. 

1.  How  much  HO,  S  O3  will  be  formed  from  one 
gramme  of  S  O2  ?     How  much  from  one  litre  ? 

While. 

Ba  Ol-{- HO,  S03-\-Aq  =  BiiO,SO3-^HCl-\-  Aq. 

BaO,NOs  +  HO,  S  0^  -\-  Aq  =  BaO,  SO3  + 
HGl-\-Aq. 

1.  Why  must  sulphuric  acid  or  a  soluble  sulphate  pro- 
duce a  precipitate  when  added,  in  solution,  to  the  solution 
of  any  salt  of  baryta  ? 

2.  The  precipitate  produced  by  adding  an  excess  of 
Ba  CI  to  a  solution  of  HO,  S  O3  was  collected,  and 
weighed  4.567  grammes.  How  much  sulphuric  acid* 
was  present  in  solution  ? 

*  When  the  name  sulphuric  acid  is  used,  H  0,  S  O3  is  always  meant,  unless 
otherwise  specified. 


69  ACIDS. 

3.  The  precipitate  produced  by  adding  an  excess  of 
HO,  S  O3  to  a  solution  of  Ba  O,  N  O5  weighed  5.942 
grammes.  How  much  Ba  0,  N  O5  did  the  solution  con- 
tain? 

172.  1st  stage.  S  +  O2={SO2,andK0,N05  +  2(ir0,  ^Og) 

=  KO,SO,.  HO,  SO,  +  MO,  IVOs. 

2d  stage.  SO2  +  HO,  ]¥05  =  HO,  S  0,  +  ]\04. 
3d  stage.  3]^04  +  ccHO=2(HO,]\05)  +  J\02. 

r    l¥02  +  02   =  ]^®4• 
4th  stage.  <  2  SO,  +  2  (H  O,  ]¥  O5)  =  2  (^0,  ^  0,) 
(      +2WO4. 

The  last  two  stages  are  now  repeated  indefinitely,  so 
long  as  there  is  a  supply  of  sulphurous  acid,  oxygen,  and 
steam,  with  the  same  amount  of  N  O4 . 

1.  How  much  sulphuric  acid  may  be  made  by  the  above 
process  from  100  kilogrammes  of  sulphur  ?  How  many 
litres  of  acid  having  the  Sp.  Gr.  1.842  ?  How  many  of 
acid  of  Sp.  Gr.  1.734?  How  many  cubic  metres  of  oxy- 
gen must  be  used  in  the  process  ?  How  many  cubic 
metres  of  air  must  pass  through  the  lead  chamber,  sup- 
posing all  its  oxygen  to  be  removed  ? 

173.  a.  HO,  SOs  -\- Ati  =  H 0,  S  0-\-  Aq, 

1.  How  much  water  must  one  kilogramme  of  the  mono- 
hydrated  acid  withdraw  from  the  air  in  order  to  reduce  its 
Sp.  Gr.  to  1.398  ? 

/.  Na  0,  CO^  -\-H0,S03-{-Aq  =  Na0,S0,  +  Aq 

^+C02. 

1.  How  much  HO,  S  O3  is  required  to  exactly  neu- 
tralize 5.645  grammes  of  anhydrous  carbonate  of  soda? 
How  much  acid  of  the  Sp.  Gr.  1.306  ? 


ACIDS.  69 

2.  How  much  NaO,  CO2  must  be  dissolved  in  one 
litre  of  water  so  as  to  make  a  solution  such  that  one  cubic 
centimetre  will  exactly  neutralize  0.01  of  a  gramme  of 
H  0,  S  O3  ? 

Yellow.  White. 

^.  Pb  O  +  HO,  SO^-\-Aq  =  PbO,  S  O3  +  Aq. 

Black.  Blue  Solution. 

h.  QxxQ  -^  H  0,  S  0^^  Aq  =  0u0,8  0^-\-  Aq, 
which,  when  evaporated,  gives  crystals  of  the  composition 

Blue  Vitriol. 

CuO,  SO3.5HO. 

1.  How  much  sulphate  of  lead  can  be  made  from  twenty 
grammes  of  litharge  ?  How  much  sulphuric  acid  must  be 
used  in  the  process  ? 

2.  How  much  crystallized  Blue  Vitriol  can  be  made 
from  one  kilogramme  of  oxide  of  copper  ?  How  much 
sulphui'ic  acid  of  Sp.  Gr.  1.615  must  be  measured  out  for 
the  process  ? 

174.  Cu  +  2  {HO,  SOs)  =  CuO,  SO3  +  2  HO  +  SOa- 

175.  C-\-2(HO,SOs)  =  2HO  +  2S02  +  C02. 
WaO,  GO,-\-SO^-}-Aq  =  JVaO,SO^-\-Aq-{-€Oi. 

1.  How  much  sulphurous  acid  can  be  made  from  4.562 
grammes  of  sulphuric  acid  by  means  of  copper  ?  How 
much  by  means  of  charcoal  ?  How  much  anhydrous  sul- 
phate of  copper  would  be  formed  in  the  first  case  ?  How 
much  carbonic  acid  in  the  second  ?  How  much  carbonate 
of  soda  will  the  sulphurous  acid  in  the  two  examples  neu- 
tralize ?  What  is  the  volume  of  S  O2,  and  what  the 
volume  of  C  O2  evolved  in  the  second  case  ? 

2.  How  much  copper  and  how  much  sulphuric  acid 
must  be  used  to  make  one  litre  of  sulphurous  acid  gas  ? 


70  ACIDS. 

How  much  to  make  500  grammes  of  anhydrous  sulphite 
of  soda? 

3.  By  burning  sulphur  in  one  litre  of  oxygen,  how  much 
S  O2  gas  is  obtained  ?     What  is  the  Sp.  Gr.  of  S  O2  ? 

Phosphorus  and  Oxygen, 
Phosphoric  Acid  (POs). 
176.  P  +  5  O  =  P  O5  (white  powder). 

Colorless  Fluid. 

3P  +  6  {HO,  NO,)  +  Aq  =  ZHO,  ,PQ,-\-  ^^ 

+  5]\02. 

For  preparation  of  phosphoric  acid  from  bones,  see  §  144. 

At  the  ordinary  temperature. 

3CaO,  ePOs  +  2(110,  SOs)-\-aq=  2(CaO,  S  O3) 
+  [2^0,  Ca  0]  ,P  O5  +  aq. 

Intensely  heated. 

2(CaO,  S03)+  [2H0,  CaOjePOs  =  SCaOePOs 

4-2(HO,  SO3). 


Before  ijcnition. 


3  HO,  P  O5  +  3  ilNH,-]  0,II0)-\-3  (Ag  0,NOs)  +  Aq 

Yellow. 

=  3AgO,  ,P05  +  3  ([i^^4]  0,  NO,)  +  Aq. 

After  ignition. 

HO,PO,-^  INH,-]  0,  HO  +  AgO,  NO,-\-Aq  = 

White. 

Ag  0,  ,P  O5  +  [^^J  0,N0,  +  Aq. 

3H0,  ,P0,  +  2{MgO,  SO,)  +  3  [iV^^^]  0,  HO 
-\-Aq  ==  [N  H4]  0,  2  Mg  0,  eP  O5  .  12  H  O  + 
2(lNH,^0,S0s)+Aq. 


ACIDS.  71 

[N  H4]  0,  2  Mg  0,  eP  Og  .  12  H  0  when  ignited  resolves 
into  2  Mg  O,  bP  O5  +  W  H3  +  13  H  O. 

1.  How  much  P  Og  and  how  much  3  H  0,  cP  Og  can  be 
obtained  from  16  grammes  of  phosphorus  ? 

2.  By  boiling  one  gramme  of  phosphorus  in  nitric  acid 
until  it  dissolves,  diluting,  neutralizing  with  aqua  ammonia, 
and  precipitating  with  a  solution  of  sulphate  of  magnesia, 
collecting  and  igniting  the  precipitate,  how  much  wiU  it  be 
found  to  weigh  ? 

3.  How  much  nitrate  of  silver  is  required  to  precipitate 
the  phosphoric  acid  made  from  one  gramme  of  phos- 
phorus before  ignition  ?     How  much  after  ignition  ? 

Cyanogen  and  Oxygen. 

179.  Cy  O    =  Cyanic  Acid,  which  is  mo7iobasic. 
Cy2  O2  =  Fulminic  Acid,  which  is  hihasic. 
Cys  O3  =  Cyanuric  Acid,  which  is  trihasic. 

Boron  and  Oxygen. 
Boracic  Acid  (B  O3). 

180.  NaO,    2BO3    +    HCl   +    Aq  =.   2  {H  0,  B  0^ 

+  iVa  CT  +  Aq. 

The  crystallized  boracic  acid  is  H  O,  B  O3  .  2  H  O. 
Dried  at  100°  it  becomes  H  O,  2  B  O3  .  2  H  O. 

At  a  red  heat  it  loses  its  water  and  melts,  and  on  cooling 
it  hardens  to  a  vitreous  mass. 


72  ACIDS. 

Silicon  and  Oxygen. 
Silicic  Acid  (Si  O3). 
Na  0,  Si  O3  -{-HCl  +Aq  =  HO,  Si  0^  -\-  Na  01 

-\-Aq. 

If  the  quantity  of  water  is  large,  the  hydrated  silicic 
acid  remains  in  solution.  If  the  amount  of  water  is  small, 
it  separates  as  gelatinous  precipitate. 


SECOND  GROUP  :  HYDROGEN  ACIDS,  OR  COMPOUNDS  OF 
THE  HALOGENS  WITH  HYDROGEN. 

Chlorine  and  Hydrogen. 
Chlorohydric  Acid  (H  CI). 

185.  Na  CI  +  HO,  S  Os  =  m  0,  SO^-^  EL  CI;  or 

NaCl  +  2  (HO,  SO,)   =   Na  0,  S 0,  .  HO,  SO, 
+  HC1. 

Only  one  equivalent  0^  HO,  S  0,  is  necessary  to  de- 
compose one  equivalent  of  salt ;  but  then  the  last  half  of 
the  H  CI  can  be  driven  off  only  at  a  temperature  suffi- 
ciently high  to  melt  glass,  so  that  with  these  proportions 
the  process  cannot  be  conducted  in  glass  vessels.  If  two 
equivalents  of  HO,  S  0,  are  used,  the  whole  of  the  H  CI 
is  expeUed  at  a  moderate  temperature,  and  the  process  can 
then  be  conducted  to  its  end  in  a  glass  flask  or  retort. 
Hence,  in  the  manufactories,  where  the  acid  is  generally 
generated  in  iron  retorts,  only  one  equivalent  of  HO,  S  O3 
is  used,  while  in  the  laboratory,  where  glass  vessels  are 
employed  in  the  process,  two  equivalents  are  taken  to 


ACIDS.  73 

each  equivalent  of  salt.     The  last  we  wiU  assume  to  be 
the  case  in  the  following  problems. 

1.  How  much  chlorohydric  acid  gas  can  be  made  from 
4.562  grammes;  from  25  kilogrammes;  from  34.567 
grammes  of  common  salt  ? 

2.  How  much  sulphuric  acid  is  required  to  decompose 
the  above  amounts  of  common  salt  ?  and  how  much  bisul- 
phate  of  soda  is  in  each  case  formed  ? 

3.  How  many  cubic  centimetres  of  H  CI  can  be  made 
from  1  gramme ;  from  5,643  grammes  of  Na  CI  ? 

4.  How  much  salt  and  how  much  sulphuric  acid  are 
required  in  order  to  make  one  kilogramme,  to  make 
5.463  grammes,  and  to  make  one  litre,  of  ffl  CI  ? 

5.  How  much  H  CI  is  contained  in  one  litre  of  the 
liquid  acid  of  Sp.  Gr.  1.16,  of  Sp.  Gr.  1.17,  of  Sp.  Gr. 
1.14? 

6.  How  many  cubic  centimetres  of  gas  are  dissolved  in 
one  litre  of  the  liquid  acid  of  the  above  strengths  ? 

7.  How  much  Na  CI,  and  how  much  HO,  SO^,  and  how 
much  water  in  the  receiver,  are  required  to  make,  — 

a.  20  kilogrammes  of  liquid  acid  of  Sp.  Gr.  1.13  ? 
h.  560.4  grammes  of  liquid  acid  of  Sp.  Gr.  1.18  ? 
c.  4  litres  of  liquid  acid  of  Sp.  Gr.  1.16  ? 

H  +  CI  =  M  CI. 

1.  One  litre  of  hydrogen  gas  combines  with  what  vol- 
ume of  chlorine  gas  ?  and  what  is  the  volume  of  hydro- 
chloric acid  gas  formed  ? 

186.  a.YQ^HCl-\-Aq  =  Fe  01 -{-Aq^  H. 

1.  How  much  liquid  acid,  by  weight  and  by  measure, 
of  Sp.  Gr.  1,16  is  required  to  dissolve  250  grammes  of 

7 


74  ACIDS. 

iron  ?     How  much  chloride  of  iron  would  be  obtained, 
and  how  much  hydrogen  gas,  by  measure,  evolved  ? 

^n  -^  H  CI  -\-  Aq  =  Sn  Gl  ^  Aq-^  H. 

1.  Solve  the  last  problem,  substituting  tin  for  iron. 

h.  FegOs,  3  HO  H-  3 ^C;  +  ^^  =  Fe^  Ck  +  Aq. 

c.  2Fem-\-  Cl^Aq=  Fe^  Ck  +  Aq. 

d.NaO,G02-\-HCl-\-Aq  =  NaCl-\-Aq-\-€Oi. 

e.  AgO,NO^-^ HCl-\-Aq  =  AgC\-\- HO, NOs-^Aq. 

1.  How  much  liquid  acid  of  Sp.  Gr.  1.16  is  required 
to  dissolve  4  grammes  of  iron-rust? 

2.  How  many  cubic  centimetres  of  chlorine  are  required 
to  convert  one  gramme  of  protochloride  of  iron  into  ses- 
quichloride  ? 

3.  How  much  liquid  acid  of  Sp.  Gr.  1.13  is  required 
to  neutralize  one  gramme  of  carbonate  of  soda  ? 

4.  How  much  H  CI  do  50  c.  c.  of  a  liquid  contain 
which  is  exactly  neutralized  by  one  gramme  of  anhydrous 
carbonate  of  soda  ? 

5.  How  much  H  CI  do  50  c.  c.  of  a  liquid  contain 
which  gives,  with  an  excess  of  nitrate  of  silver,  a  precipi- 
tate weighing  5.643  grammes  ? 

Aqua  Eegia. 

H0,N0,-\-ZHCl-\-aq  =  N02Ck-^Cl^  aq. 
Au  +  {N  0^  Ck  -\-0l-\-  aq)  =  Au  Ck  +  aq  +  N  O^. 


ACIDS.  75 

Bromohydric  Acid  (HBr).      lodohydric  Acid  (HI). 

PBrs  +  SiTO  =  P03  +  3HBr. 
PI3  _^  3  iTO  =  PO3  +  3  H  I. 

Hydrofluoric  Acid  (H  Fl). 

CaFl  4-  RO,SO-\-  aq  =  CaO,  SO3  +  HFl-\-aq. 
Si03  +  3HFl  =  3HO  +  SiFl3. 

Tartaric  Acid  (2  H  0,  Cs  H4  0,o)  • 

194.  2  UN  H,-]  0,  HO)   +  ^  H  0,   O,  H,  0,,  -^  Aq  =^ 
^INH,-]  0,  GsJT,0,o  +  Aq. 

2  (K  0,    CO,)    +   2  ^  0,    Gs  IT,  O.o    -}-    Aq   = 
2K0,  Gsffi  0,0  +  ^?  +  2  C  O2. 

2K0,  GsJI,  0,0  +  HGl  +  ^^  =  HO,  KO,  Cs H4O10* 

-^KGl-\-Aq. 

2  (CaO,  0  00  +  2  (irO,  KO,  Gs  H,  0,,)  -\-  Aq  = 
2  CaO,  C4 HsO.o  +  2K0,GsH,  0,,  +  ^^  +  2  CO2. 

2/f  0,  Gs  H^  6>,o  -^2GaGl^Aq=2  Ca  0,  Cg  H4  0,o 

^2KGl^Aq. 

2  CaO,  C4  HsOio  +  2  {HO,  S0-^^Aq==2  (Ca  0,  S  O3) 
^2 HO,  G,HsO,,^Aq. 

*  This  salt  is  not  absolutelj^  insoluble,  but  only  difficultly  soluble  in  -water, 
and  hence  is  not  completelj'-  deposited  in  this  reaction. 


76  ACIDS. 

1.  How  much  KO,  CO2  is  required  to  exactly  neu- 
tralize 5.462  grammes  of  tartaric  acid  ? 

2.  How  much  H  CI  is  required  to  convert  4.678  grammes 
of  2  K  O,  Cs  H,  0,0  into  H  O,  K  O,  C3  H4  Oio  ? 

3.  From  ten  kilogrammes  of  cream  of  tartar  how  much 
tartaric  acid  can  be  made  ? 

OxaKc  Acid  (H  0,  C2O3). 

The  above  is  the  sypabol  of  the  acid  dried  at  100°. 
When  crystallized,  it  contains  two  more  equivalents  of 
water,  and  corresponds  to  the  formula  H  0,  C2  O3  .  2  H  O. 

196.  H  O,  Ca  O3  .  2  ^0  +  ^  {HO,  SOs)  =  x  {HO,  SO,) 

197.  h.KO,CO.,-{-  HO,  G^  0,  +  Aq  =  KO,  G^  0,  +  Aq 

+  C  O2. 

KO,  0,  O3  +  HO,  C^O,  +  Aq  =  KO,^  a  6)3  +  Aq. 

d.  Ca  0,  SO,  ^  HO,   Oi  0,  -\-  Aq  =  Ca  O,  C2O3 

-\-HO,SO,  +  Aq. 

Ca  0,SOs^  iNH,-]  0,  G,  0,  ^  Aq  =  Ca  0,  C2  O3 
+  {NH,)  0,S0,-^  Aq. 

1.  How  much  C  O2  and  how  much  C  O  will  be  ob- 
tained by  decomposing  five  grammes  of  crystallized  oxalic 
acid  by  sulphuric  acid  ?  How  many  cubic  centimetres  of 
each  gas  ? 

2.  How  much  crystallized  oxalic  acid  must  be  used  to 
yield  one  litre  of  C  O2  and  one  litre  of  C  O  ? 

3.  How  much  crystallized  oxalic  acid  will  exactly  neu- 
tralize 1.456  grammes  of  carbonate  of  potassa  ? 


ACIDS.  77 

Acetic  Acid  {HO,  [C4  ^3]  O3). 
198.  PbO  +  HO,  \_G^H,-]  0^  -\-  Aq  =  Pb  0,  [O.Hs]  0, 

■^Aq. 

CrystaUized  acetate  of  lead  =  Pb  O,  [Q  H3]  O3 .  3  H  0. 

PhO,  (GiHs)  Os  +  HO,  SOs  +  Aq  =  PbO,  SO3 
-\-HO,lC,H,-]  0,-\-Aq. 


7* 


LIGHT    METALS. 


FIKST    GROUP  :    ALKALI   METALS. 

Potassium  (K). 

202.  KO,  aO^-\-HO,iG,H,-]  0,-]- Aq  =  KO,\_G,H,-\  0, 

+  Jl^  4-  C  02. 

KO,  CO2+  HO,  S03  +  ^^=XC>,  ^^(93  +  ^g+COj. 

1.  How  much  HO,  S  O3  must  be  dissolved  in  water 
in  order  to  make  a  litre  of  test  acid  such  that  one  cubic 
centimetre  will  exactly  neutralize  one  decigramme  of 
KO,  CO2? 

2.  How  much  crystallized  oxalic  acid  must  be  dissolved 
in  water  in  order  to  make  a  litre  of  test  acid  such  that 
one  hundred  cubic  centimetres  will  exactly  neutralize  6.92 
grammes  of  K  O,  C  O2  ? 

203.  K  0,   G  O2  -\-    Ga  0,  H  0  -\-  Aq   =    Ca  O,  C  O2 

-^KO,  HO  -\-  Aq. 

Melted  together. 

204.  d.SiOs-\-KO,I£0=  K 0,  Si  O3  +  H ©. 

Light  Blue. 

e.  Gic  0,  S  O3  -\-  K  0,  If  0  -\-  Aq  ===   Cu  O,  H  0 

+  KO,SO,  +  Aq. 


LIGHT   METALS,  79 

Intensely  heated. 

205.  KO,  C O2  -(-  2  C  =  K  +  3  C  O. 

206.  KO,  00^  +  2  {HO,  SO^) -{-aq  =  KO,  SO3.  HO,  S O3 

+  a^  +  C02. 

207.  KO,  G0,-\-  HO,  N 0^  -\-  aq  =  KO,  N  O5  -{-  aq 

+  CO2. 

When  heated. 

a.  KO,  NO5  =  KO,  NO3  +  O2. 

J.  K  0,  NOg  +  3  C  +  S  =  K  S  +  3  €  O2  +  TV. 

1.  How  many  cubic  centimetres  of  mixed  gases  are 
formed  by  the  burning  of  one  kilogramme  of  gunpowder, 
when  measured  at  the  standard  temperature  and  pressure  ? 
Assuming  that  the  temperature  at  the  time  is  1000°,  what 
would  be  the  volume  of  the  gases  the  moment  after  the 
explosion. 

2.  Assuming  that  gunpowder  occupies  the  same  bulk  as 
an  equal  weight  of  water,  into  how  many  times  its  own 
volume  does  it  expand  on  burning  ?  Calculate  both  for 
0°  and  for  1000°. 

3.  Assuming  that  the  temperature,  the  moment  after 
explosion,  is  1000°,  what  would  be  the  pressure  on  the  in- 
terior surface  of  a  bomb  of  20  centimetres  internal  diame- 
ter when  exploded  filled  with  gunpowder  ? 

208.  a.  K  O,  CI  O5  =  K  CI  +  6  O. 

1.  How  much  does  the  oxygen  contained  in  chlorate  of 
potassa  expand  when  the  salt  is  decomposed  ?  Sp,  Gr. 
of  chlorate  of  potassa  is  2  nearly. 

2.  How  much  mechanical  force  would  be  required  to 
reduce  oxygen  gas  to  the  same  degree  of  condensation  in 
which  it  exists  in  the  salt  ? 

c.  3  (KO,  CI  0,)  +  4  {HO,  ^  O3)  =  2  (K  0,  SO3 .  HO,  SO3) 

+  K0,  C10;  +  2ClO4. 


80  LIGHT   METALS. 

f. -KO,  C\Os+  ^ H Gl -^r  Aq  ==  KGl -{-  Aq  -\-  2 CI. 
6  {KO,  HO)  -\- <o  CI  -^  Aq  = -KO,  QIO, -Y  6  K 01 

+  Aq. 

210.  KI  ^  MnO,  +  2{H0,  S  0^)  -\-  aq  =  K  0,  S  Os 

-^  Mn  0,  S  Os -\- aq -{-  I. 

211.  H  O,  K  0,  Cs  H4  Oio  =  Cream  of  Tartar. 
Na  0,  K  O,  Cs  H4  Oio         =  Eochelle  Salts. 

[N  Hi]  0,  K  0,t)8  Hi  Oio  =  Ammoniated  Tartar. 

Fea  O3,  K  0,  Cg  H4  Oio       =  Tartarized  Iron 

Sbij  O3,  K  0,  Cs  H4  Oio       =  Tartar  Emetic. 

B  O3  K  0,  Cs  Hi  Oio  =  Soluble  Cream  of  Tartar. 

2 13.  4  (K  0,  C  O2)  +  1 6  S  =  K  0,  S  O3  +  3  K  Ss  +  4  C  ©8 . 

3(K0,  C02)  +  8S  =  KO,  S2O2+2KS34-3CO2. 

The  first  or  the  last  of  these  reactions  takes  place, 
according  to  the  proportion  of  sulphur  and  the  degree 
of  temperature  to  wliich  the  mixture  is  exposed.  If  the 
sulphur  is  in  excess,  and  the  temperature  of  the  mass 
raised  to  a  red  heat,  pentasulphuret  of  potassium  and  sul- 
phate of  potassa  are  formed.  If,  however,  the  sulphur  is 
present  in  smaller  quantity,  and  the  temperature  of  the 
mass  is  not  raised  above  its  melting  point,  a  mixture  of 
tersulphide  of  potassium  and  hyposulphite  of  potassa  re- 
sults. At  the  higher  temperature,  hyposulphite  of  potassa, 
which  is  always  first  formed,  splits  up  into  sulphate  of 
potassa  and  pentasulphide  of  potassium.     Thus, 

4  (K  0,  S2  O2)  =  3  K  O,  S  O3  +  K  Ss . 

Either  of  these  sulphides  is  decomposed  by  dilute  acids, 
forming  sulphide  of  hydrogen,  and  setting  free  as  many 
equivalents  of  sulphur,  less  one,  as  are  contained  in  the 
compound. 


LIGHT   METALS.  81 

KO,  S  O3  +  4  C  =  K  S  +  4  C  O. 

Sodium  (Na). 

215.  Problems  on  Common  Salt. 

1.  One  gramme  of  salt  contains  how  much  chlorine  and 
how  much  sodium  ? 

2.  One  cubic  centimetre  of  common  salt,  Sp.  Gr.  = 
2.13,  contains  how  many  cubic  centimetres  of  sodium,  and 
how  many  of  chlorine  gas  ? 

218.  Na  CI  +  ^(9,  /S  O3  =  Na  0,  SO3  +  H  CI. 

219.  NaO,S  03  +  40  =  NaS  +  4CO. 

220.  Na  S  +  CaO,  C  O2  =  Ca  S  +  Na  O,  C  Oj. 

3  (NaO,  S O3)  +  13  C  +  4  (CaO,  CO^)  =  3 NaO, COa') 
+  3  Ca  S,  Ca  0  +  14  C  O. 

1.  How  much  carbonate  of  soda  can  be  made  from  500 
kilogrammes  of  common  salt  ?  How  much  sulphuric  acid  ? 
How  much  charcoal  and  how  much  chalk  are  required  in 
the  process,  according  to  the  theory  ? 

2.  If  in  a  chemical  process  carbonate  of  potassa  or  car- 
bonate of  soda  may  be  used  indifferently,  which  would  be 
employed  most  profitably  if  the  price  were  the  same  ? 

3.  "What  relation  ought  the  price  of  crystallized  car- 
bonate of  soda  to  bear  to  that  of  the  dry  salt,  if  the  intrin- 
sic value  is  alone  considered  ? 

221.  Na  0,    CO,   -{-   Ca  0,  H 0  -\-   Aq   =   Ca  0,   0  0^ 

•\-NaO,HO.i-\-Aq. 


82  LIGHT    METALS. 

222.  Na  0,  C  Oa  +  2  C  =  Na  +  3  C  O. 

223.  2  {Na  0,  GO^)-\-n HO,PO,  +  Aq  ^ 

HO,  2mO,POs-\-Aq^2€  O2. 

The    symbol    of    crystallized    phosphate    of    soda  = 
HO,  2NaO,  PO;.24HO. 

When  heated. 

HO,  2NaO,  POg.24HO=2NaO,  PO5  +  25HO. 

3(Ag  0,  NO,)   +  HO,   2  Na  0,    P  0,  -^  Aq  = 

Yellow. 

3 KgO,  V0,-{- 2  {Na0,N0s)-\~H0,N0s  +  Aq. 

While. 

2  {Ag  0,  NO,)  +  2iVa  0,  P  O5  +  ^^  =  2  AgO,  P  O5 
-{- 2  {Na  0,  N  Os)  +  Aq. 

1.  How  much  sodium  can  be  made  from  one  kilogramme 
of  anhydrous  carbonate  of  soda  ? 

2.  How  much  Ag  O,  N  O5  is  required  to  precipitate 
completely  the  phosphoric  acid  from  1.345  grammes  of 
crystallized  phosphate  of  soda  ?  How  much,  from  the 
same  amount  of  pyrophosphate  of  soda  ? 

224.  Na O,  N  O5  +  2  {HO,  S 0,)  =  Na  0,  S  0„  HO,  SO, 

+  MO,  ]\05. 

NaO,NO,^KGl-\-aq  =  '^2LQ\-\-KO,NO,-^aq. 

225.  Na  O,  2  B  O3 .  10  H  O  =  Symbol  of  common  borax. 
NaO,  2BO3.    5H0=      «       "     octohedral  borax. 

1.  "When  soda  nitre  and  potash  nitre  bear  both  the  same 
price,  from  which  can  nitric  acid  be  most  profitably  ex- 
tracted ? 

2.  When  potash  nitre  is  worth  25  cents  the  kilogramme, 
and  chloride  of  potassium  8  cents  the  kilogramme,  at  what 


LIGHT    METALS.  83 

price  of  soda  nitre  will  it  be  profitable  to  convert  it  into 
potash  nitre,  assuming  that  the  cost  of  the  process  amounts 
to  two  cents  on  each  kilogramme  of  potash  nitre  manu- 
factured ? 

3.  What  is  the  percentage  composition  of  common 
borax  ?     What  is  that  of  octohedral  borax  ? 

4.  How  much  borax  glass  can  be  made  from  100 
grammes  of  common  borax  ? 

Ammonia  ([N  HJ  0). 

The  above  is  the  symbol  of  the  base  of  the  ammonia 
salts,  and  corresponds  to  K  O  and  Na  O.  The  student 
'must  be  careful  not  to  confound  it  with  ammonia  gas, 
which  has  the  symbol  N  H3 . 

227.  K  O,  H  0  +  Fe  =  K  O  +  Fe  O  +  M. 

K  O,  N  O5  -f  5  Fe  =  K  O  +  5  Fe  O  +  J¥. 

3  (KO,  H  0)  -I-  K  0,  N  O5  +  8  Fe  =  4  K  0  +  8  Fe  O 

229.  J\  H3  +  II  CI  =  [N  HJ  CI. 

[NHJCl  +  CaO,  H0=  Ca  CI  +  2HO  +  ]¥Il3. 

230.  ^^  +  M  Ms  =  INff.l  0,H0^  'Aq. 

231.  INH,]  0,H0  +  Aq-\-2U.^=lNH,-]S,HS-^Aq. 

The  compound  which  remains  in  solution  after  pass- 
ing H  S  gas  through  aqua  ammonia  until  saturation,  is 
[N  H4]  S,  H  S,  the  sulphohydrate'  of  sulphide  of  ammo- 
nium, which  is  the  reagent  so  much  used  in  the  laboratory, 
and  incorrectly  called  sulphide  of  ammonium.  On  ex- 
posure to  the  air,  the  solution  becomes  yellow,  owing  to 


84  LIGHT    METALS. 

the  formation  of  bisulphide  of  ammonium,  as  is  shown  in 
the  reaction  below. 

\_NH,-\  S,  HS  +  [_NH,-\  0,H0-{-  Aq  = 
2iNH^-]S-\-Aq. 

Colorless  Solution. 

2  (IN  H,-]  S,  H  S)  +  Aq   -]-  ^  O    = 

Yellow  Solution. 

lNH,-\  S,  +  IFH,-]  0,  S,  0,  +  Aq.        ' 

232.  2  [N  H4]  CI  +  3  (Ca  O,  C  0^)  =  3  Ca  CI  +  JV  H3 
+  2  []V  M  J  ®,  3  C  O2  +  H  O. 

1.  How  many  cubic  centimetres  do  17  grammes  of 
ammonia  gas  occupy?  How  many  do  36.5  grammes  of 
chlorohydric  acid  gas  occupy  ?  "When  the  two  gases  com- 
bine, in  what  proportions,  by  volume,  do  they  unite  ?  How 
great  is  the  condensation  which  results  ?  Sp.  Gr.  of 
[N  H4]  Ci  =  1.5. 

2.  How  much  ammonia  gas  can  be  obtained  from  5 
grammes  of  chloride  of  ammonium.  How  much  chloride 
of  ammonium  and  how  much  Ca  O  must  be  used  in  order 
to  prepare  one  litre  of  ammonia  gas  ? 

3.  How  much  chloride  of  ammonium  and  how  much 
lime  must  be  used  in  order, to  prepare  one  litre  of  aqua 
ammonia  of  Sp.  Gr.  =  0.9  ?  How  much  water  must  be 
placed  in  the  receiver  ? 

4.  How  much  ammonia  gas  is  held  in  solution  by  one 
litre  of  aqua  ammonia  of  Sp.  Gr.  0.9  ?  To  how  much 
[N  H43  O,  H  O  does  this  amount  of  gas  correspond  ? 

5.  How  much  H  0,  S  O3  is  required  to  neutralize  four 
cubic  centimetres  t»f  aqua  ammonia  of  Sp.  Gr.  0.9  ? 

6.  How  much  H  S  gas  will  be  absorbed  by  one  litre 
of  aqua  ammonia  of  Sp.  Gr.  0.9,  assuming  that  only  so 
much  is  taken  up  as  is  necessary  to  form  the  compound 


LIGHT   METALS.  85 

[N  H4]  S,  H  S  ?  How  much  H  0,  S  O3  and  Fe  S  will  be 
required  to  produce  this  amount  of  gas  ?  How  much 
additional  aqua  ammonia  must  be  added  to  the  above 
solution  in  order  to  change  it  to  a  solution  of  proto- 
sulphide  of  ammonium  ? 


SECOND    GROUP  :   THE   ALKALINE   EARTHS. 

Calcium  (Ca). 

241.  CaO,SO^-{-BaCl^Aq=B2^0,^03-\-CaCl-\-Aq. 

Ga  0,  S  0,  -\-  H  0,    G^  0^  +   ^^   =   Ca  0,  C^  O3 
-]-HO,SO,-\-Aq. 

242.  HO,2N'aO,PO,-\-2GaGl-\-Aq=ILO,2Q2iO,VO, 

+  2  iVa  a  +  Aq. 

244.  2  (Ca  0,  H  O)  +  2  CI  =  Ca  0,  CI  O  +  Ca  CI. 
246.  Ca O,  C O2  +  ^CT  +  ^^  =  Ga  Gl -\- Aq -{-  CO2. 

1.  How  much  Ca  O  can  be  obtained  from  100  kilo- 
grammes of  carbonate  of  lime  ?  How  much  Ca  0,  H  0 
can  be  obtained  from  the  same  amount  ? 

2.  How  much  carbonate  of  lime  must  be  burnt  in  order 
to  yield  140  kilogrammes  of  quicklime  ?  How  much  to 
yield  185  kilogrammes  of  Ca  O,  H  O  ?  How  many  cubic 
metres  of  C  0^  would  be  set  free  during  the  process  ? 

3.  How  many  cubic  metres  of  C  O2  can  be  absorbed  by 
a  quantity  of  milk  of  lime  containing  5  kilogrammes  of 
CaO? 

4.  "What  is  the  percentage  composition  of  unburnt  and 
of  burnt  gypsum  ? 


86  LIGHT   METALS. 

5.  What  is  the  percentage  composition  of  phosphate  of 
lime  (3  Ca  0,  P  O5)  ? 

6.  How  much  chlorine  and  how  much  lime  are  re- 
quired to  make  100  kilogrammes  of  chloride  of  lime, 
assuming  that  its  composition  is  expressed  by  the  symbol 
Ca  O,  CI  O  +  Ca  CI  ?  How  much  Mn  O2  and  how  much 
muriatic  acid  of  Sp.  Gr.  1.15  will  yield  the  requisite 
amount  of  chlorine  ? 

7.  To  how  much  Ca  O,  how  much  Ca  CI,  and  how  much 
Ca  O,  S  O3  do  2.5  grammes  of  carbonate  of  lime  correspond? 

Barium  and  Strontium  (Ba  and  Sr). 

248.  Ba  O,  S  O3  +  4  C  =  Ba  S  +  4  C  €». 

Ba  S  -{-  H  CI  -^  Aq  =  Ba  CI  -\-  Aq  -\-m^. 

QnO  +  Ba  S -\-  Aq  =  CnS  -\-  Ba  0,  H 0  -\-  Aq. 

Ba  CI  -\-  NaO,  S  0,  -\-  Aq  =  Ba  0,  S  O3  +  Na  CI 
+  Aq. 

Sr  O,  S  O3  +  4  C  =  Sr  S  +  4  C  ©. 

SrS-^HO,  NO,  J\-Aq  =  SrO,NO,-\-Aq-\-m^. 

Sr  0,  N  0,  +   Na  0,  S  0,    +    Aq   =    Sr  0,  S  O3 
NaO,NOs-\-Aq. 

1.  What  is  the  percentage  composition  of  sulphate  of 
baryta  ? 

2.  How  much  chloride  of  barium  can  be  made  from  5 
kilogrammes  of  sulphate  of  baryta  ?  How  much  nitrate 
of  baryta,  and  how  much  Ba  O,  can  be  made  from  the 
same  amount  of  sulphate  of  baryta  ? 

3.  To  how  much  sulphuric  acid  do  4.567  grammes  of 
sulphate  of  baryta  correspond  ?  To  how  much  chloride 
of  barium  do  they  correspond  ? 


LIGHT   METALS.  87 

4.  How  much  nitrate  of  str-ontia  can  be  made  from  one 
kilogramme  of  sulphate  of  strontia  ?  How  much  carbon 
and  how  much  nitric  acid  of  Sp.  Gr.  1.4  is  required  in 
the  process  ? 

Magnesium  (Mg). 

249.       The  symbol  of  serpentine  mineral  is 

9[Mg,Fe]0,  4Si03,  6H0. 

The  symbol  [Mg,  Fe]  O  indicates  that  a  portion  of  the 
magnesium  iu  the  base  is  replaced  by  oxide  of  iron,  and 
the  whole  stands  for  but  one  equivalent  of  base.  We 
frequently  write,  instead  of  the  portion  enclosed  in  brack- 
ets, the  general  symbol  R,  when  the  symbol  of  the 
mixed  base  becomes  E.  O,  and  that  of  serpentine  mineral 
9  E  O,  4  Si  O3.  It  must  be  noticed,  that  Mg  and  Fe  when 
enclosed  in  brackets,  with  a  comma  between,  as  above,  no 
longer  stand  for  an  equivalent  of  each  metal.  On  the 
other  hand,  the  two  together  make  but  one  equivalent  of 
metal,  represented  by  R  in  the  other  mode  of  writing. 
Moreover,  nothing  is  intended  to  be  indicated  in  regard  to 
the  relative  proportions  of  the  two  metals  mixed  together 
to  form  one  equivalent,  as  they  vary  in  different  speci- 
mens of  the  same  mineral.  This  method  of  writing  the 
symbols  of  compounds  containing  isomorphous  constitu- 
ents Avill  be  constantly  used  hereafter. 

9  [Mg,  Fe]  0,  4  Si  O3 .  6  H  O  +  9  (^6>,  ^  0^)  -\- Aq  ^ 
4  Si  O3  +  9  [j%,  Fe-]  0,S0,-{-  Aq. 

Since  sulphate  of  magnesia  and  sulphate  of  protoxide 
of  iron  have  the  same  crystalline  form,  we  shall  obtain,  on 
evaporating  the  solution,  crystals  containing  both  salts. 
We  can  prevent  the  sulphate  of  protoxide  of  iron  from 


88  LIGHT   METALS. 

crystallizing,  by  converting  it  into  sulphate  of  sesquioxide 
of  iron  by  means  of  nitric  acid.     Thus, 

6  {Fe  0,  aS'  O3)  +  3  {HO,  S  0,)  +  HO,  NO,  -^  Aq  ^ 
3  {Fe^  O3,  3  ^  O3)  +  ^^  +  W  ©2. 

Talc  =  6MgO,  5  SiOs,  2  H  0. 

Meerschaum  =  Mg  O,  Si  O3,  H  0. 

Hornblende   =  4  [Mg,  Ca,  Fe]  0,  3  Si  O3. 

Augite  =  3  [Ca,  Fe,  Mg]  0,  2  Si  O3. 

250.  5  {Mg  0, SO,)  +  5  {KO,  00^)  -{-Aq=d (MgO,  CO2 . aq) 

+  MgO,  HO  +  %  0,  2  (76>2  +  5  {K 0,  S 0^) -^  Aq. 

The  relative  proportions  of  carbonate  of  magnesia  and 
of  hydrate  of  magnesia  vary  with  the  temperature  and 
other  circumstances  attending  the  precipitation. 

251.  M.g0,Q0^-\-HGl^Aq  =  Mga^Aq-\-i^O2. 

2  {Mg  0,  S0s)-\-H0,2  Na  0,P0,-\-  {N H,-\  0,H0 
-{-Aq=  [NHJ  0,  2MgO,  PO5  +  2  {Na  0,  S 0,) 

JrAq. 

When  heated. 

(NH4)0,2MgO,P05=2MgO,P054-WM3  +  HO. 

1.  What  is  the  percentage  composition  of  talc  ?  What 
that  of  hornblende  and  augite,  assuming  that  the  whole  of 
the  base  in  either  case  is  Mg  O  ? 

2.  From  a  solution  of  sulphate  of  magnesia  the  whole 
of  the  magnesia  was  precipitated  by  phosphate  of  soda 
and  ammonia.  This  precipitate,  after  ignition,  was  found 
to  weigh  2.456  grammes.  How  much  sulphate  of  mag- 
nesia was  contained  in  the  solution  ? 


LIGHT   METALS.  89 

Aluminum  (Al). 

258.  AI2O3,  Si  O3 .  2  H  O  +  3  {HO,  S  O3)  -\-  aq  =  SiOs 

+  AI2O3,  3S03  +  a(?. 

AU  0^,^SO^  +  Z{N'aO,  G 0^)^Aq  =  A\0^,ZB.O 
+  3  {Na  0,  S03)+Aq-\-C  ©2. 

AI2O3,  3  HO  +  KO,  HO-\-Aq  =  KO,  Ak  O3  +  Aq. 

KO,SO^.  Ak  O3,  3  aS  C>3  +  3  {Fa  0,  O  0,)  -{- Aq  = 
AI2  O3,  3  HO  +  3  (iVa  O^S  O3)  +KO,SOs-{-  Aq 

+  C®2.  . 

Al,  Os,  3S0,  -\-  S  (Pb  0,  10,  H,-\  O3)   +  Aq  = 
3(PbO,  SO3)  +^4  0„  3  lO,H,-\  0,  +  Aq. 

Symbols  of  Isomorphous  Alums. 

Potassa,  Alumina,  Alum. 
K  0,  S  O3 .  AI2  O3,  3  S  O3 .  24  H  O. 

Soda,  Alumina,  Alum. 
NaO,  S O3 .  AI2O3,  3  S O3 .  24 HO. 

Ammonia,  Alumina,  Alum. 
[N H4] O,  SO3 .  AI3 O3,  3 SO3  .  24HO. 

Potassa,  Chrome,  Alum. 
KO,  S  O3 .  Cr2  O3,  3  S  O3 .  24  HO. 

Soda,  Chrome,  Alum. 
Na  0,  S  O3 .  Crg  O3,  3  S  O3  .  24  H  O. 

Ammonia,  Chrome,  Alum. 
[N  HJ  0,  S  O3 .  Cr2  O3,  3  S  O3 .  24HO. 


90  LIGHT   METALS. 

Potassa,  Iron,  Alum. 
K O,  S O3 .  Fea O3,  3  S  O3 .  24 HO. 

Soda,  Iron,  Alum. 
NaO,  S  O3 .  Fes  O3,  3  S  O3  .  24:HO. 

,     Ammonia,  Iron,  Alum. 
[NH4]  0,  S  O3 .  FeaOs,  3  S  O3  .  24HO. 

Symbols  of  the  most  important  Silicates  of  Alumina. 

2AI2O3,  SiOg,  Staurotide. 

3  AI2O3,  2  SiOs,  Andalusite. 

3  AI2  O3 ,  2  Si  O3 ,  Ky anite. 

3  AI2  O3,  2  Si  [O,  Fl]3,  Topaz, 

K  O,  Si  O3 .  AI2  O3,  3  Si  O3,  Common  Felspar. 

Na  O,  Si  O3 .  AI2  O3,  3  Si  O3,  Albite. 

[Ca,  Na]  0,  Si  O3  .  AI2  O3,  Si  O3,  Labradorite. 

K  0,  Si  O3 .  4  (AI2  O3,  Si  O3),  Common  Mica. 

3  R  O,*  Si  O3 .  R.  03,t  Si  O3,  Garnet. 

1.  What  is  the  percentage  composition  of  staurotide  ? 
What  is  that  of  kyanite  ? 

2.  An  analysis  of  one  of  the  above  silicates  would  give 
the  foUowing  percentage  composition. 


Silica, 

64.76 

Potassa, 

16.87 

Alumina, 

18.37 
100.00 

What  is 

the  symbol 

of  the 

mineral  ? 

*  K  0  =  Fe  0,  Mn  0,  Mg  0,  or  Ca  0. 
t  Ks  Os  =  AI2  O3,  Fe2,  O3,  or  Cra  O3. 


LIGHT   METALS.  91 


Solution.  —  This  problem  is  evidently  the  reverse  of  deducing  the  per- 
centage composition  from  the  symbol ;  but  it  does  not  admit,  like  that,  of 
a  definite  solution,  for  while  there  is  but  one  percentage  composition 
corresponding  to  a  given  symbol,  there  may  be  an  infinite  number  of 
symbols  corresponding  to  a  given  percentage  composition.  This  can 
easily  be  made  clear  by  an  example.  The  commonly  received  symbol 
of  alcohol  is  [C4  H5]  0,  H  0  =  C4  Ho  O2 .  The  percentage  composition 
is  easily  ascertained.    Thus, 

C4     Hg    O2 

24  +  6  +  16  =  46. 
46  :  24  =  100  :  a;  =  52.18  per  cent  of  carbon. 
46  :    6  =  100  :  x  =  13.04  per  cent  of  hydrogen. 
46  :  16  =  100  :  x  =  34.78  per  cent  of  oxygen. 

Percent. 

Carbon,          52.18  =  Ca  =  12  or  C4  =  24  or  Cs  =  36 

Hydrogen,      13.04  =  H3  =    3  "  Hs  =    6  "  H9  =    9 

Oxygen,         34.78  =  0  =_8  "  O2  =_16  "  O3  =  2i 

100.00                  23  46  69 

This  percentage  composition  evidently  corresponds  not  only  to  C4  Hs  O2 , 
but  also  to  C2  H3  0,  to  Cg  H9  O3 ,  and  to  any  other  symbol  -which  is  a 
multiple  of  the  first;  for,  taking  the  per  cent  of  carbon  as  an  example, 
■we  have 

100  :  52.18  =  23  :  12  =  46  :  24  =  69  :  36  =  92  :  48,  &c. 

If,  then,  we  had  given  the  percentage  composition  of  alcohol,  it  would 
be  impossible  to  determine,  without  other  data,  whether  the  symbol  was 
C2  Ha  0,  or  some  multiple  of  it.  If,  however,  we  had  also  given  that  the 
sum  of  the  equivalents  of  the  elements  of  alcohol  equalled  46,  then  we 
could  easily  reverse  the  above  process.    Thus, 

100  :  52.18  =  46  :  a;  =  24,  the  sum  of  the  equivalents  of  carbon. 
100  :  13.04  =  46  :  a;  =    6,        "  "  "  "       hydrogen. 

100  :  34.78  =  46  :  a;  =  16,       "  "  "  «       oxygen. 

—  =  4,  number  of  equivalents  of  carbon. 


hydrogen. 


—  ==  6,        "         "  " 

16 

—  =  2,        "        "         "         "      oxygen. 

Assuming,  however,  that  we  had  no  means  of  ascertaining  the  sum  of 
the  equivalents  in  alcohol,  then,  although  we  could  not  definitely  fix  its 
sjmibol,  yet  nevertheless  we  could  easily  find  which  of  all  the  possible 
symbols  expressed  its  composition  in  the  simplest  terms ;  in  other  words, 


92  LIGHT    METALS. 


■with  the  fewest  number  of  whole  equivalents.  For  this  purpose,  assume 
for  a  moment  that  the  sum  of  the  equivalents  is  equal  to  100,  then 

52.18  =  the  sum  of  the  equivalents  of  carbon. 
13.04=      "  "  "  "       hydrogen. 

34.78  =      "  "  "  "       oxygen. 

— ^  =  8.697,  number  of  equivalents  of  carbon. 

13^  =  13.04,        «  "         "  "  hydrogen. 

84  78 

— ^  =  4.348,        "  "  "  "  oxygen. 

8 

These  are  the  number  of  equivalents  of  each  element  on  the  supposition 
that  the  sum  of  the  equivalents  in  alcohol  is  equal  to  100.  Any  other 
possible  number  of  equivalents  must  be  either  a  multiple  or  a  submul- 
tiple  of  these,  and  we  can  easily  find  the  fewest  number  of  whole 
equivalents  possible,  by  seeking  for  the  three  smallest  whole  numbers 
■which  stand  to  each  other  in  the  relation  of  8.697  :  13.08  :  4.348, 

■which  will  be  found  to  be  2:3:1. 

Hence  the  simplest  possible  symbol  is  C2  H3  0,  but,  from  anything  we  are 
assumed  to  know,  the  symbol  may  be  any  multiple  of  this ;  and  for  con- 
siderations which  cannot  be  discussed  in  this  connection,  chemists  usually 
assign  to  alcohol  the  symbol  C4  He  O2 ,  which  is  double  the  above. 

The  symbol  thus  obtained  expresses  merely  the  relative  number  of 
equivalents  of  each  element  present  in  the  compound,  and  gives  no  in- 
formation in  regard  to  the  grouping  of  the  elements.  Such  symbols  are 
called  empirical  symbols,  to  distinguish  them  from  the  rational  symbols, 
which  indicate  the  manner  in  which  the  elements  are  supposed  to  be 
arranged.  The  rational  symbol  of  alcohol  is  [C4  H5]  0,  H  0.  This  in- 
dicates not  only  that  alcohol  consists  of  four  equivalents  of  carbon,  six 
of  hydrogen,  and  two  of  oxygen,  but  also  that  it  is  the  hydrated  oxide 
of  a  compound  radical  called  ethyle.  It  must  be  carefully  noticed,  how- 
ever, that  the  empirical  symbols  fully  represent  all  our  positive  knowl- 
edge. They  alone  are  not  hable  to  be  changed.  The  gi-ouping  of 
elements  in  a  compound  is  a  matter  of  theory,  and  the  rational  symbols 
are  liable  to  constant  changes,  as  the  opinions  of  chemists  on  this  sub- 
ject vary. 

From  the  example  just  discussed  we  can  easily  deduce  the  following 
rule  for  finding  the  empirical  symbol  of  a  compound  from  its  percentage 
composition.  Divide  the  per  cent  of  each  element  entering  into  the  com- 
poundhy  its  chemical  equivalent^  and  find  the  simplest  series  ofichole  num- 
bers to  ivMch  these  results  correspond.  To  apply  this  rule  to  the  problem 
under  consideration. 


LIGHT  METALS.  93 


— 1-—  =  1.43,  number  of  equivalents  of  silica. 
45.3 

—1— =  0.3575,    "       "         "       "      potassa. 
47.2 

i-— -  =  0.3575,    "       "        "       "      alumina. 
51.4 

143  :  0.3575  :  0.3575  =  4:1:1. 

Empirical  symbol,  AI2  O3 ,  K  0,  4  Si  O3 . 

Rational  symbol,  K  0,  Si  O3 ,  AI2  O3 ,  8  Si  Oa . 

3.  An  analysis  of  one  of  the  silicates  of  alumina  would 
give  the  following  percentage  composition. 

Per  cent. 

Silica,  53.29 

Lime,  16.47 

Alumina,  30.24 

100.00 

What  is  the  symbol  of  the  minei'al  ? 

Ans.  Ca  0,  Si  O3 .  A\,  O3  Si  O3. 

Second  Method  of  Solution.  —  By  inspecting  the  formula  obtained  by 
solving  the  problem  according  to  the  method  just  described,  the  student 
will  see  that  the  amount  of  oxygen  in  the  acid  stands  in  a  very  simple 
relation  to  that  in  the  bases.  This  relation  is  1  :  3  :  6,  corresponding  to 
Ca  0,  AI2  O3 ,  and  2  Si  O3 .  It  has  been  shown  in  the  text-book,  §  200, 
that  a  similar  simple  ratio  exists  between  the  amount  of  oxygen  in  the 
acid  and  that  in  the  bases  of  all  oxygen  salts.  The  I'atio  can  easUy  be 
found  from  the  percentage  composition.  For  this  purpose  we  have 
merely  to  calculate  the  amount  of  oxygen  in  the  per  cent  of  the  acid  and 
bases  indicated  by  analysis,  and  find  the  simplest  ratio  in  which  these 
amounts  stand  to  each  other.    In  our  example, 

53.29  per  cent  of  silica  contains  28.24  parts  of  oxygen. 
16.47     "        "      lime         "  4.71      "      "      " 

30.24      "        "     alumina  "        14.12      "      "      " 

According  to  the  principle  just  stated,  these  numbers  ought  to  stand  to 
each  other  in  some  simple  ratio,  and  it  can  easily  be  seen  that 

4.71  :  14.12  :  28.24  =  1:3:6. 

From  this  ratio  we  can  easily  deduce  the  symbol,  for  one  equivalent  of 
oxygen  corresponds  to  one  equivalent  of  Ca  0,  three  equivalents  of  oxy- 
gen correspond  to  one  equivalent  of  AI2  O3 ,  and  sis  of  oxygen  to  two 


94  LIGHT   METALS. 

equivalents  of  Si  O3 .  Hence  the  symbol  is  Ca  0,  AI2  O3 ,  2  Si  O3 ,  which 
we  may  write  as  above,  Ca  0,  Si  O3  .  AI2  O3 ,  Si  O3.  For  convenience 
in  calculating  the  amount  of  oxygen  from  the  per  cent  of  acids  or  bases 
indicated  by  analysis,  Table  VII.  has  been  added  at  the  end  of  the  book, 
which  gives  the  per  cent  of  oxygen,  together  with  its  logarithm  con- 
tained in  the  bases  and  acids  of  most  common  occurrence. 

In  deducing  empirical  symbols  from  the  results  of  actual  analysis,  it 
must  be  remembered  that  our  processes  are  not  absolutely  accurate,  and 
that  therefore  we  must  not  expect  to  find  more  than  a  close  approxima- 
tion to  a  simple  ratio  between  the  oxj-gen  in  the  base  and  that  in  the 
acid.  Again,  in  mineral  compounds,  it  is  very  frequently  the  case  that 
isomorphous  bases  replace  each  other  to  a  greater  or  less  extent.  This  is 
the  case  in  common  garnet,  the  symbol  of  which  may  be  written  thus : 

8  [Fe,  Jin,  Mg,  Ca]  0,  Si  O3  .  [Al2Fe2]  O3,  Si  O3. 
We  generally,  however,  write  the  symbol  as  on  page  90 : 

3  R  0,  Si  O3  .  E2  O3 ,  Si  O3 . 
Here  E  0  stands  for  the  sum  of  all  the  protoxide  bases,  which  make 
together  but  one  equivalent  of  base,  and  E2  O3  for  the  sum  of  all  the 
sesquioxide  bases,  which  also  make  together  but  one  equivalent  of  base. 
Such  general  symbols  as  these  give  all  the  information  in  regard  to  the 
constitution  of  the  mineral  which  is  required.  In  deducing  such  sym- 
bols, it  is  evident  that  the  oxygen  of  all  the  protoxide  bases  must  be 
added  together  to  obtain  the  amount  of  oxygen  in  the  assumed  base 
E  0,  and  aU  the  oxygen  of  the  sesquioxide  bases  must  be  added  together 
in  the  same  way  in  order  to  obtain  the  amount  of  oxygen  in  the  assumed 
base  E2  O3 .  From  these  sums  we  can  easily  obtain  the  required  oxygen 
ratio. 

4.  An  analysis  of  andesine  (a  mineral  allied  to  felspar) 
yielded  the  following  result. 


Proportion  of  Oxyg'en. 

Silicic  Acid, 

59.60 

30.90  in  Si  0. 

Alumina, 
Sesquioxide 

of  Iron, 

24.28 
1.58 

":S}="-™""^o 

Lime, 

5.77 

1.61- 

Magnesia, 

1.08 

0.37 

-  =  3.79  in  RO. 

Soda, 

6.53 

1.65 

Potassa, 

1.08 

0.16. 

99.92 

What  is  the  symbol  of  the  mineral  ? 


LIGHT   METALS.  95 

Solution.  —  The  ratio  of  the  oxygen  in  E  0,  R2  O3,  and  Si  O3  is 
3.79  :  11.70  :  30.90  =  1  :  3.08  :  8.1, 
for  which  we  may  substitute,  for  reasons  stated  above, 
1:3:8. 
One  equivalent  of  oxygen  corresponds  to  R  0.  • 

Three  equivalents  of  oxygen  correspond  to  Eg  O3 . 
Eight  equivalents  of  oxygen  correspond  to  ^  Si  O3  • 
But  as  we  do  not  admit  fractional  equivalents,  we  may  multiply  the 
whole  by  three,  when  we  obtain  the  empirical  symbol 

3E0,  3E2  03,8Si03; 
from  which  we  may  deduce  the  rational  symbol 
3  E  0,  2  Si  O3  .  3  (E2  03,  2  Si  O3). 

5.  Deduce  the  symbols  of  the  silicious  minerals  of  which 
the  following  are  analyses. 

1. 
Silicic  Acid,  65.72 

Sesquioxide  of  Iron, 
Alumina,  18.57 

Lime,  0.34 

Magnesia,  0.10 

Potassa,  14.02 

Soda,  1.25 

100.00       100.0       100.04       100.16 


2. 

68.4 

3, 

44.12 

63.70 

0.1 

0.70 

0.50 

20.8 

35.12 

23.95 

0.2 

19.02 

2.05 

0.66 

0.65 

0.25 

1.20 

10.5 

0.27 

8.11 

HEAVY    METALS. 


FIRST   GROUP    OF   THE    HEAVY   METALS. 

Iron  (Fe). 
286.  a.   4  Fe  +  O  =  Fe4  O. 

b.  3  Fe4  O  +  13  O  =  4  (Fe  0,  Fe^  O3). 

c.  2  (Fe  0,  Fe.2  O3)  +  O  =  3  Fe^  O3. 

d.  2  (Fe  O,  S  O3.   6  H  O)  when  heated  resolves  into 

Fg  O3,  S  O3  +  S  O2  + 12  H O ;  by  further  heat- 
ing, Fe^  O3,  S  O3  =  Fea  O3  +  S  O3. 

e.  3  Fe  +  4  O  =  Fe  0,'Fe2  O3. 

/.    2  (Fe  0,  Fea  O3)  H-  9  ^  0  +  ©  =  3  (Fe^  oTs  H  0). 
g.  Fe   O,    Fes  O3  -\-  2   G  0^   +  Aq  =  Fe^  O3 
^Fe  0,2  O  0,-\-Aq. 

Red. 

2  (Fe  0,  2  0  O2)  +  ^^  +  -O  =  Fe^  O3,  3  H  0 
^4,0  0,-^  Aq. 
285.        Fe  +  IlO,S03-{-Aq  =  FeO,SO,-j-Aq-{-n^ 


HEAVY    METALS.  97 

285.        Fe  +  Cii  0,  ^  O3  +  ^^  =  Cu  +  Fe  0,  S  0, 

Red. 

a.   6  (Fe  0,  S  0,) -\- Aq  +  3  O  =  ¥e,  O3,  3  H  0 

Yellow-Brown. 

-{-2(Fe^0s,  3  S  0,)+Aq. 
h.    6  {Fe  O,  S  0,)  +  3{H0,  S  0,)  -{- H  0,  N  0^ 

Yellow-Brown. 

■\-Aq=3  {Fe^  Os,  3  S  O3)  +  .4^  +  ]¥  O2 . 

Light  Green. 

c.  FeO,SOs-\-lIfJI,2  0,ffO-^Aq  =  FeO,B.O 

+  [iV^^4]  0,  SOs-\-Aq. 
Fe^  Os,  3SOs  +  3  ([JSTIf,]  0,  H  0)  -\- Aq  = 

Red. 

Fe2  O3,  3  H  0  +  3  ([J^IIi']  0,  S  0^)  +  Aq. 

d.  '2^YQ-}-A:HO,NO,-^rAq  =  Fe^Os,3NO,-\-Aq 

288.        \_H0,  2  Na  0]  P  Os  +  3  {Fe  0,  S  0^)  -\- Aq  = 

White. 

3  Fe  O,  V  0,^2Na  0,  SO^^HO,  SOs 

~\-Aq. 

[HO,  2NaO']POs-\-  Fe^  Os,  3  S  Os  +  Aq  = 

While. 

Fe^Os  ,P05.4HO+2  {M  0,  SOs)-\-HO,  SO^+Aq. 

290.  3  Fe  O    +    2  Fea  O3    +    ^  H  Cy  -^  Aq   = 

Blue. 

3  Fe  Cy,  2  Fca  €73  +  9  HO  +  Aq. 

Blue. 

291.  3  Fe  Cy,  2  Fcg  Cyg  +  %  {K  0,  HO)  +  Aq  = 

,   Red. 

2  (Fe2  O3,  3  H  0)  +  3  (2  ir  (7y,  Fe  Cy)  +  Aq. 

292.  a.   3  (2  ^  (7y,  Fe  Cy)  +  2  (^eg  Os,  3  SOs)+Aq  = 

3  Fe  Cy^^'Fea  Cyg  +  6  (.ff  0,  >«?  O3)  +  Aq,. 

9 


98  HEAVY  METALS. 

292.  5.    2  ^  Gy,    Fe  Cy  ^  2  {Fe  0,    S  0,)   -^  Aq  = 

White. 

3  Fe  Cy  4-  2  Z"  0,  5  O3  +  Aq. 

White.  Blue. 

9  Fe  Cy  +  3  O  =  Fe,  O3  +  3  Fe  Cy,  2  Fez  Cyg. 
c.   2  K  Cy,  Fe  Cy  -\-  2  {Cu  0,   S  O3)   -\-  Aq  = 

Purple. 

2  Cu  Cy,  Fe  Cy.  Aq  +  2  {KO,  S  0,)  +  Aq. 
2  Z-  (7y,  Fe  Cy  -i-  2  (Pb  0,  ]^  0,)  -\-  Aq  = 

White. 

2  Pb  Cy,  Fe  Cy.  Aq  +  2  (JT  0,  iV^Os)  +  Aq. 
A  large  number  of  similar  compounds  having  the 

general  symbol  2  R  Cy,   Fe  Cy.    Aq,  may  be 
formed  thus :  — 

White.  Pale  Yellow. 

2  H  Cy,  Fe  Cy.  Aq.  2  [N  H/]  Cy,  Fe  Cy.  Aq. 

Yellow.  Yellow. 

2  Na  Cy,  Fe  Cy.  Aq.       2  Ba  Cy,  Fe  Cy.  Aq. 

Pale  Yellow.  White. 

2  Mg  Cy,  Fe  Cy.  Aq.       2  Zn  Cy,  Fe  Cy.  Aq. 

There  are  also  compounds  in  which  the  two  equiva- 
lents of  R  in  the  general  symbol  are  replaced  by 
different  metals  thus  :  — 

White.  Yellow. 

Yellow  Salt.  Red  Salt. 

293.  2  (2  X  Cy,  Fe  Cy)^  Cl-{- Aq  =Z  K  Cy,  Fe^  Cy, 

-^KCl-\-  Aq. 

3  K  Cy,  Fe^  Cy,  +  3  {Fe  0,    S  0,)  -\-  Aq  = 

3  Fe  Cy,'"Fe2  Cys  +  3  {K  0,  S  0,)  +  Aq. 


HEAVY   METALS.  99 

293.  There  also  may  be  formed  a  large  nmnber  of 
similar  compounds  having  the  general  symbol 
3  E.  Cy,  Fcg  Cys,  such  as :  — 

Brown.  Lijht-Red. 

8  H  Cy,  Fe,  Cy,.  8  Ca  Oy,  Fe^  Cyj. 

{^B^Cyl-'^'^Cys.  Aq. 

Light-Gtecn  Solution.  Black. 

294  Fe  0,  S  0,  +  \_N  H,-\  ^  +  ^^  =  Fe  S 
+  IN-H,-]  0,SO,-\-Aq. 

Black.  Lig-ht-Green. 

295.        Fe  S  +  a^  +  4  O  =  J^e  0,  ^  O3  +  a^. 


Manganese  (Mn). 

299.  Mn  O2  +  iT  0,  S  0^ -\- aq  =  Mn  0,  S03-\-aq 

+  0. 

Black.  Lijht-Pink  Solution. 

MnO^-^^  2  H  Gl  -]-  Aq  =  Mn  01  -^  Aq  -\-  CI. 

Light-Pink  Solution.  Brown. 

300.  a.   6  {Mn  0,  /S  O3)  +  .ig  +  3  O  =  Muj  O3,  3  H  O 

+  2  {Mn^  Os,  S  S  Os)+  Aq. 

Light-Pink  Solution.  'White. 

b.    MnO,S03-{-l]^iri']0,H:0-\-Aq  =  MnO,'H.O 

-\-[i^ir,-]o,so,  +  Aq. 

White.  Brown. 

2  (Mn  0,  H  0)  +  «?  +  O  =  Muj  O3,  3  H  O 
-\-aq. 

Lig-ht-Pink  Solution.  Flesh-colored. 

Mn   0,    S  Os  -\-   [iV^  ff,-]  S  -\-  Aq  =   Mn  S 
-{-[JSTH,-]  0,SO,-i-Aq. 

Melted  together. 

301.  MnO^-f  K  0,11 0-^0=  K0,Mn03  +  UO. 


100  HEAVY  METALS. 


Green  Solution, 

301.        3  {K  0,  Mn  O3)  +  ^y  +  2  C  O2  =  Mn  O2 

Crimson  Solution. 

-^KO,  Mn^  Ot-\-2{KO,G  0^)  +  Aq. 

Green  Solution. 

3  {K  0,  Mn  Os)  -}-  2  (JIO,   S  O3)  -{-  Aq  = 

Crimson  Solution. 

Mn  Oi+KO,  Mn^  Ot^2{KO,S  0,)J^Aq. 


Cobalt  (Co)  +  Nickel  (Ni). 

Pink  Solution.  Blaclc. 

303.        Co  0,    S  Os  -{-  [N-  H,-\   S  -[-  Aq  =   Qo  ^ 


\_NH,-]  0,SO,  +  Aq. 

Green  Solution. 

+  [i\rir,]  o,sOs  +  Aq. 


Green  Solution.  Black. 

M   0,   S  Os  -^  [JSr  H,-]   S  -{-  Aq  =  m  ^ 


Zinc  (Zn). 

311.  Zn  -{-  H  0,  S  0^  -{-  Aq  =  Za  0,  S  Os  -\-  Aq 

+  H. 

White. 

312.  a.   Zn  0,  S  Os  -\-  KO,  H  0-{-  Aq  =^ZnO,  no 

■\-KO,SOs-\-  Aq. 
This  precipitate  dissolves  in  an  excess  of  K 0,  HO 
+  Aq. 

White. 

b.    2k  0,    S  Os    -i-    [li  H,-]  S  -\-  Aq  =   Zn  ^ 
^iNH,-\0,SOs  +  Aq. 


HEAVY    METALS.  ^01, 

312.  c.    5  {Zn  0,  S  0,)  -^  6  {Na  0,  G  0,)  -{-  Aq  = 

While.  White, 

2    (Zn    0,     C    Oa)     +     3    (Zn   O,     H   0) 
+  5  (iVa  0,  S  0^)  -\-  Aq  -\-  3  €  O,. 
This,  precipitate  is  a  mixture  of  Zn  O,  C  O2,  and 
Zn  O,  H  O,  but  in  variable  proportions. 


Yellow. 

315.        Cd  0,S  Os  +  JIS-\-Aq=CdS-\-irO,S03 


Dark-Brown.  Gray-White.  White. 

317.        The  oxides  of  tin  are  Sn  O  ;  Sng  O3  ;  and  Sn  Oj. 


Cadmium  (Cd). 

a 

+  Aq. 

Tin  (Sn). 

Dark-Bi 

tin  are  Sn 

319.  Sn  -^  H  Gl ^  aq  =  Sn  01  -\-  aq  -{-  n. 

White. 

320.  Sn  Gl  +  \_NH^  0,  H  0  -{-  Aq  =  SnO,  HO 

+  [iV^^4]  Gl^Aq. 

321.  Sn  Gl -\-  Gl -\- Aq  =  Sn  Gl,  +  Aq. 

Sn -\- 2  I£  Gl  -i-  IT  0,  W  Os  -\-  Aq  =  Sn  CI, 

-\-Aq  +  NOs. 

White. 

SnGk-\-2  ilJ^Ifi']  0,irO)^Aq  =  lIO,  Sn 0^ 

+  2  [i^^,]  Gl-\-Aq. 
SnO+KO,HO-^Aq  =  KO,Sn  0 -\- Aq.. 
HO,  Sn  O2  +  ^  0,  HO  -\-Aq  =  KO,  Sn  Q, 

+  Aq. 

9* 


102  HEAVY  METALS. 

By  evaporation  of  the  last  solution  we  can  obtain 

crystals  of  K  O,  Sn  Oj.   4  H  O. 

We  may  also  prepare  Na  O,  Sn  Og .   4  H  O. 

White. 

S24.        5  Sn  +  10  {H  0,  N 0,)  -\- Aq  =  Sng  0,o.  10  H  O 
^Aq -\-10NOi. 
Sng  Oio.     10  H  O    +    K  0,   H  0  -\-  Aq  = 

KOySn,  0,,^Aq. 
By  evaporation  we   can   obtain  crystals  of 

White. 

K  0,  Sng  Oio.   4  H  O. 

Brown. 

325.        Sn  CI  -{-  ITS  -{-  Aq  =^  Sxi  S  -}-  H  Gl  -{-  Aq. 

Yellow. 

Sn  Ck  +  2  HS-\-Aq  =  Sn  ^^ -\- ^  H  Gl -\- Aq. 


SECOND  GROUP  OF  THE  HEAVY  METALS. 

Lead  (Pb). 

Black.       Red  or  Yellow.    Red-Yellow. 

331.        The  oxides  of  lead  are  Pbj  O ;    Pb  O ;    Pba  O3 ; 

Dark-BrowD, 

and  Pb  Oj. 
334,        B  Vh -\- 4.  {H  O,  N  0,) -\- aq  =  B  (Pb  0,  W  0^) 

+  a^  +  ]¥  O2. 
3  Pb  0-1-  3  (^0,  iV^  O5)  +  a^  =  3  (Pb  0,  N  0,) 
-\-  aq. 


HEAVY    METALS.  103 

White. 

335.  Ph  0,  N  Os  +  HO,  S  0^-\-Aq  =  Pb  O,  S  O3 

+  HO,NOs  +  Aq. 
Ak  O3,  3  >Sf  (93  +  3  {Pb  0,  [C;  ^3]  O3)  +Aq  = 

White. 

3  (Pb  O,  S  O3)  +  Ak  O3,  3  [Ci  H,-]  O3  +  Aq. 

White. 

336.  Pb  O  +  ^  (7?  +  a^  =  Pb  CI  +  aq. 
Vh  O  +  H  Gl -\- Aq  =  Pb  CI  +  Aq. 

337.  Acetate  of  oxide  of  lead  =  Pb  O,  [C^  H3]  O3 .  3  H  O. 
Basic  acetate  of  oxide  of  lead  = 

3  PbO,  [QHsjOa.  HO. 

338.  2  (Pb  0,  [  C4  ^3]  6>3)  +  2  IT  0,  0^  B,  0,,  -\-Aq^ 

White. 

2  Pb  O,  Cs  H,  Oio  +  2  (IT  (9,  [ Oi  H,-\  0,).+  Aq. 

White. 

PbO,N'0,-\-lNH,-\  0,HO-\-Aq  =  VhO,B.O 
JrWH,-]0,  NO,-^Aq. 

339.  Pb    0,    [C,  Bs]    Os    +    Aq    4-    2   Pb  O   = 

3Pb  0,lG,ff,-]  0,  +  Aq. 
3  Pb  0,    IC  ffs]    O3   +   ^?  +  2  C  O2  = 

While. 

2  (Pb  O,  C  O2)  +  P&  0,  [a  ^3]  O3  +  Aq. 

340.  Zn    -f    P5    (9,      [C;   Hs]    O3    -{-    Aq    ==    Pb 

+  ^«  6>,  [en,-]  Os-i-Aq. 

Blaclc. 

341.  P^  0,    [O;  1^3]   O3  +  ^  .S'  +  ^^  =   Pb  s 

+  ^0,  [(7,^3]  0,-\-Aq. 

342.  Pb  S  +  3  O   =  Pb  O  +  S  O2,    also  Pb  S 

White. 

+  4O  =  Pb0,  SO3. 


104  HEAYT  METALS. 

342.  By  roasting  galena  we  obtain  a  mixture  of  Pb  O 
with  a  small  amount  of  Pb  O,  S  O3.  By  then 
melting  together  Pb  O  or  Pb  0,  S  O3  and  an 
excess  of  Pb  S,  we  obtain  metallic  lead  and  sul- 
phurous acid,  thus  :  — 
2  Pb  O  +  Pb  S  =  3  Pb  +  S  O2,  also  Pb  0,  S  O3 
+  PbS  =  2Pb  +  2S02. 

Bismuth  (Bi). 

847.        2  Bi  +  4  (^  (9,  II  O5)  +  «^  =  Bi^  O3,  3  J^  0^ 
-\-aq-\-N  O2. 

White. 

4  (Bi,  Oe,    SI^Os)  -\-  Ag  =  3  (Bi^  O3,   N  O5) 

+  Bh  O3,  9  i\r  O5  +  Aq. 

Brownish-Black. 

Bh   O3,    d  If  Os  -\-  3  II  S  -{-  Aq  =  Bk  S3 
-\- 9  (HO,  If  0,)  +  Aq. 

Copper  (Cu). 

Green. 

349.  Malachite  =  Ou  0,  HO.  Cu  O,  CO2. 

Blue    Carbonate   of   Copper   = 

Blue. 

CuO,  HO.  2  CuO,  CO2. 

Red.  Black. 

350.  The  oxides  of  copper  are  Cu2  0  and  Cu  O. 

Blue  Solution.  Blue. 

352.         Cu  0,  S  0,  +  K  0,  H  0  -\-  Aq  =  Cvi  O,  n  O 

■  -\-K0,  SO,-\-Aq. 


HEAVY  METALS.  l05 

Blue.  Black. 

352.  By  boiling,   Cu  0,  B.  0 -{- Aq  ^  Cn  O -}-  Aq. 

Light-Blue  Solution. 

353.  GuO,    ^03  +  2  ilNH,-]  0,  H  0)  -^  Aq  =^ 

Very  deep  Blue  Solution. 

Oit    0,    S   Os-    2  IT  Bs.     H  0  -i-  Aq   = 

[^^l]  0,  S  0,.  [_NH,-\  0,H0^Aq,2. 

solution  which,  when  treated  with  alcohol,  yields 
crystals  having  the  composition 

Deep  Blue. 

JN^^I  0,   SO3.   [NHJO,  HO. 
These,  when  heated,  are  resolved  into 

Green  Powder. 

|n  H^3  I  0,  S  O3  +  []¥  H4]  O,  H  O. 

854.  2((7mO,  >S03)  +  2(^0,  ir(9)+^g  — 0*  = 

Yellow-Red. 

Cu2  O  +  2  {K  0,  S  O3)  +  Aq. 

Heated  together. 

855.  2  (Cu  0,  S  O3)  +  2  (Na  O,  C  O2)  -f  C  =  2  Cu 

+  2  (Na  O,  S  O3)  +  3  C  O2* 

856.  Zn  +  Gu  0,  S  O3 -\- Aq  =  Cu  ^  Zn  0,  S  O3 

+  Aq. 

357.        Cu  0,  H  O  +  H  =  Cu  +  2  H  O. 

Black.  Green  Solution. 

359.  CnO-^HOl-\-Aq=  Cu  Cl  +  Aq. 

Blue  Solution. 

360.  3  Cu  +  4  (^  (9,  i\r Og)  +  «y  =  3  (Cic  0,  JSf  0,) 

+  a^  +  TV  O2. 


*  The  oxygen  is  removed  by  adding  grape  sugar  to  the  solution. 


106  HEAVY  METALS. 

Blue  Salt.  Black.  White. 

360.  2(CuO,  NO5.  4H0) -f  Sn=2CuO  +  Sn02 

2  ((7m  0,   S  O3)  -{-  2  {Fa  0,  G  0,)  -\-  Aq  = 

Blue  or  Green. 

Cu  O,   H  O.   Cu  0,  C  O2  +  2  {Na  0,   S'O,) 

+  ^?  +  co,. 

3  (Cu  0,  S  0,)  -^IHO,  2 Na  0],  POs  +  Aq== 

Greenish-Blue. 

3  Cu  O,  P  O5  +  2  {Na  0,  S  0,)  +  H  0,  S  0, 

JrAq. 

Emerald-Green. 

Dioptase  =  3  Cu  0,  2  Si  O3.   3  H  0. 

361.  Basic  Acetate  of   Copper   = 

Green. 

Cu  0,  [C4  H3]  O3.   Cu  O,  H  O.  5  H  O. 

Neutral  Acetate  of  Copper  = 

Green. 

CuOCaHsiOg.  5  HO. 

Blue  Solution.  Black. 

362.  GuO,SO^-^rHS-\-Aq=Cvi^-\-HO,SO^ 

-\-Aq. 

Black.  Green  Solution. 

CxxS^HGl-{-ciq=  Ou  Gl-i-aq-^JS.^. 

Mercury  (Hg). 

367.  Q  Bff -{- A  (ff  0,  N  Os) -i- aq  =  S  (ffff^  0,  N  Os) 

+  «^  +  ]¥  O2. 

Black. 

368.  Hff^  0,  NO,  -{-  K  0,   H  0  -\-  Aq  ==  Hg^  0 

J^KO,  N0,  +  Aq. 


HEAVY  METALS.  107 

368.  Hahnemann's  Suboxide  of  Mercury  has  not  a 
constant  composition.  Some  chemists,  however, 
assign  to  it  the  symbol   2  Hga  0,  N  O5.   N  H3. 

White. 

370.  Hg^    0,    N  0^  -\-  Na    CI  ^  Aq   =   Hg^   CI 

-\-Na  0,  N  0,-[-  Aq. 

371.  d  Hg-^4.{H0,N  0,)-\-aq  =  d{Hg  0,N  0,) 

+  aq-\-N  O,. 

Yellow. 

ITg  0,    NO,  -{-  KO,    IT  0  -{-  Aq   =   Hg  0 

+  K  O,  N  O5  +  Aq. 

When  heated.  Red. 

372.  Hg  O,  N  O5  =  Hg  O  +  ]¥  O4  +  O. 

373.  ffg  0,  N  Os -{-  Na  Gl  -\-  Aq,  gives  no  precipitate, 

because  Hg  CI  is  soluble  in  water. 

White. 

B.g  O -\- IT  CI -\- aq  =  Ug  CI -{- IT  0 -{- aq. 

Heated  together.  Residue.  Sublimate. 

Hg  O,  S  O3  +  Na  CI  =  Na  O,  S  O3  +  Hg  CI. 

Black. 

Hg2  CI  +  ^  0,  ITO  -\-  Aq==  Hg^O-^-K  CI 

+  Aq. 

Yellow. 

Hg  CI  +  .AT  0,  i^  0  -i-  ^^  =  Hg  O  +  ^  (7/ 

+  Aq. 

White. 

374.  2  Bg  Gl-\-  [iV^iT,]  0,  HO-\-Aq  =  {n^^  X  ci 

-\-Ha-\-Aq. 

Black  Powder. 

375.  Hg  Cl-^  Sn  Gl-\-Aq  =  Hg  +  Sn  Ck  +  Aq. 

Black. 

376.  HgCl  +  HS-^-Aq^-S-g^^  HCl-[-Aq. 


108  HEAVY  METALS. 

Silver  (Ag). 

380.  3  Ag  +  4  {HO,  NO,)-{-Aq  =  Z  {Ag  0,  NO,) 

+  ^^  +  W  O2. 

Heated  together. 

381.  a.   Ag  O,  N  O5  +  2  C  =  Ag  +  2  C  Oa  +  W  O2. 

Brown, 

h.   AgO,NO,-\-KO,irO+Aq  =  AgO,E.O 
-i-KO,  NOs  +  Aq. 

White. 

Ago,  nO-\-lNH,-\  0,  HO  +  Aq==  {^Ag} 
+  Aq. 

White. 

c.    Ag    0,    N  0,    -^    Na    Gl   -\-    Aq  ^   Ag   CI 
+  iVa  6>,  A^  O5  +  Aq. 

Black 

e.   AgO,  NO,-\-HS-\-Aq^Ag^-\-HO,  N  0, 

-{-Aq, 

Gold  (Au). 

385.        Axi-\-ZHGl  +  HO,  NO^  ■{-  Aq=Au  Ok 
+  ^g  +  W  O2. 

Dark-Brown, 

387.  Au  Ok  +  6  {Fe  0,  S  O3)  +  .4^  =  Au  -f  Fe^  Ok 

+  2{Fe,  O3,  3^03)+^^. 

388.  Aurate  of  Potassa  (K  O,  Au  O3  +  4  H  0)  is  a 

compound  of  oxide  of  potassium  and  teroxide 
of  gold,  in  which  the  last  plays  the  part  of  an 
acid. 


HEAVY   METALS.  109 

Platinum  (Ft). 

391,  BVi  -^  ^HGl  -\-  2{H0,  N  0^)  -{■  Aq  = 

3  P<  C4  +  ^g  +  2  W  O2. 

Yellow. 

392.  lNH,-\  CI -\-PtCk^-Aq=  [N  HJ  CI,  Pt  Cl^ 

+  Aq. 

Yellow. 

394,        K  01  +  Pt  Ok  -\-Aq=K  CI,  Pt  CI2  +  Aq. 

Black. 

The  Oxides  of  Platinum  are  Pt  0*  and  Pt  Oj. 

Greenish-Brown.    Reddish-Brown. 

The  Chlorides  of  Platinum  are  Pt  CI  and  Pt  CI2. 

Black.  Black. 

The  Sulphides  of  Platinum  are  Pt  S  and  Pt  Sj. 

Black. 

Pt  01  +  H  S  -^  Aq  =  -2tS  -{-  H  01  ^  Aq. 

Black, 

Pt  (74  +  2  ir5+  ^^  =  Pt  Sa  +  2  iTCZ  +  Aq. 


THIRD  GROUP  OF  THE  HEAVY  METALS. 

Chromium  (Cr). 
397.        The  symbol  of  chrome  iron  ore  is  Fe  O,  Crg  O3 ; 
but  almost  invariably  a  portion  of  the  Fe  O  is 
replaced  by  Mg  0,  and  a  portion  of  the  Crj  O3 


Heated  togrether  in  contact  with  air. 

2  (Fe  0,  Cra  O3)  +  2  (K  O,  C  0^)   +  7  O  = 


by  AI2  O3. 

Heated  t 

(Fe  0,  Cra 

Fe^  O3  +  2  (K  O,  2  Cr  O3).  +  2  C  O2 

*  Only  known  in  combination  with  water. 
10 


110  HEAVY   METALS. 

397.  The  above  process  is  hastened  by  mixing  with  the 
pulverized  mineral  a  portion  of  nitre,  which  yields 
when  heated  a  large  supply  of  oxygen. 

Red.  Yellow. 

898.  K0,2Cr  0^-^K0,  O0^^Aq=2  {KO,  Cr  O3) 
+  4?  +  C  O2. 

Yellow.  Red. 

2  (KO,  Cr  0,)-\-HO,NOs  +  Aq=KO,  2  Cr  O3 
J^KO,  N0,-\-  Aq. 

399.  K  0,    Cr  O3  +  Pb  0,   IC,  H,']   0^  -\-  Aq  =. 

Yellow. 

Pb  O,  Cr  O3  +  ^  0,  [C;  ^3]  O.^Aq. 

Yellow.  Red. 

-  2  (PbO,  Cr03)  -^KO,  HO-\-Aq  =  2  Pb  0,  Cr  O3 
+  K0,  Cr  6>3  +  Aq. 

Yellow.  White. 

400.  2  (Pb  O,   CvO,)  -^  S  H  CI  -\-  Aq  =  2  Pb  CI 

Green. 

+  Cr^  Ck  +  .4^  +  3  CI. 

Cr^    Ck  +    3    {IN  H,-\    0,    H  0)   +  Aq   = 
Cra  O3,  10  H  O  +  3  [Nil,-]   CI  +  Aq. 

KO,  2  Cr  Os-\-  H  0,  S  0,-\-  ^  S  C,-\-  Aq  = 
KO,  S  0,.   Cr^  6>3,  3  ^  O3  +  ^^. 

The   symbol   of    crystallized   chrome   alum   is 
KO,  SO3.   Cr^Os,  3SO3.   24  HO. 

401.  KO,  2  Cr  Os  -\-  X*  {JI  0,  S0,)-\-aq=2  CrOg 

-\-  KO,  SO^.  H 0,  S  Os  +  X  {IT  0,  S  0,) 
-\-aq. 

*  X  is  here  used  to  express  an  indefinite  amount. 


HEAVY    METALS.  Ill 

Red.  Green. 

401.  a.    2  Cr  O3  —  3  O  =  Cv^  O3. 

The  oxygen  in  the  last  reaction  may  be  removed  by 
alcohol  or  any  other  reducing  agent. 

Antimony  (Sb). 
The  oxides  of  antimony  are  as  follows  :  — 

White. 

403.  Oxide  of  Antimony,    Sb  O3. 

White. 

Antimonious  Acid,   Sb  O4  =  ^  (Sb  O3,  Sb  O5). 

Pale-Yellow. 

Antimonic  Acid,   Sb  O5. 

404.  3Sb+4(jy(9,i\^C»5)=3Sb04+4HO+4]\O2. 
When  antimony  is  treated  with  an  excess  of  con- 
centrated nitric  acid,  only  Sb  O4  appears  to  be 
formed.  If,  however,  the  acid  is  dilute,  the  anti- 
monious acid  is  mixed  with  more  or  less  of  basic 
nitrate  of  antimony  (2  Sb  O3,  N  O5)  according  to 
the  degree  of  dilution. 

By  heating  together  one  part  of  metallic  antimony 
and  four  parts  of  nitre  in  a  crucible,  there  is 
formed  a  white  mass,  which  is  a  mixture  of  anti- 
moniate  of  potassa  (K  0,  Sb  O5)  ;  nitrite  of 
potassa  (K  O,  N  O3)  ;  and  undecomposed  nitre 
(K  O,  N  O5).  Warm  water  will  dissolve  the 
two  last,  but  not  the  antimoniate  of  potassa.     If, 


112  HEAVY  METALS. 

however,  this  anhydrous  salt  is  boiled  with  water 
for  one  or  two  hours,  it  combines  with  five  equiv- 
alents of  water,  forming  a  soluble  compound 
(K  O,  Sb  O5 .  5  H  0).  The  white  mass,  which 
seemed  at  first  insoluble,  dissolves  in  great  meas- 
ure, leaving  in  suspension  only  a  small  amount  of 
binantimoniate  of  potassa. 

405.  ^hS3-{-3irCl-^aq=SbCls-{-aq-}-Sn^. 
Sb  +  3  ^  CZ  +  (II  0,   N  0,)  -^aq^Sb  Ck 

+  ag  +  ]V  O2. 

Sb  +  X  CI  =  ^5  (74  +  X  CI. 

Sh  Ck^-  Aq  =  ^h  O^^  B  H  CI  -\-  Aq. 

The  precipitate  which  is  first  formed  on  diluting  a 
concentrated  solution  of  Sb  CI3  with  water,  always 
contains  some  chloride  with  the  oxide ;  but  by  con- 
tinued washing  with  water,  or  still  better,  with  a 
weak  solution  of  Na  0,  C  Og,  the  whole  will  be 
converted  into  oxide. 

Sh  Gk  +  ^^  =  Sb  O5  +  5  IT  CZ  +  Aq. 

406.  Sb  O3  4-   [^  0,    K  0]   a  H,  0,,  +  Aq  = 

IKO,  ShO,-]  G,H,  0,,-\-Aq. 
The  symbol   of   crystallized  tartar   emetic  is 
[K  0,  Sb  O3]  Cs  H4  Oio.  2  H  O. 


HEAVY    METALS.  113 

407.  IK  0,  Sh  Os]  OsH,  0,0  +  S  JiS-\-Aq  =  Sb'Sj 

[IT  0,  K  0]  G,  H,  0:o  +  Aq. 
Kermes   mineral  is  an  amorphous  modification  of 

Sb    Sg. 

Bri^ht-Ycllow. 

Sb  Ol,-{-5  ffS-\-Aq=  Sh^,-{-5IICl-{-Aq. 
Golden  Sulphuret  is  a  mixture  of  Sb  S3  and  Sb  S5 . 

Melted  together. 

408.  Sb  S3  -j-  3  Fe  =  Sb  +  3  Fe  S. 


Arsenic  (As). 

412.  As  -j-  3  O  =  As  O3  (arsenious  acid). 

413.  2  As  O3  +  3  C  =  2  As  +  3  C  O2. 

414.  As  O3 -\- K  0,  IT  0 -\- Aq  =  K  0,  As  O3 -}-  Aq. 
a.  K  0,    As   0,  -\-  2  (Ou   0,    S  O3)   -{-  Aq  = 

2  Cu  OrAs  O3  +  ^  0,  SO,.  BO,  S  0,-{-  Aq. 
■    h.    The   symbol  of    Schweinfurth   green  is 
Cu  O,  [Q  H3]  O3.   2  Cu  O,  As  O3. 

415.  As  O3  +  2  {H  0,    N  0,)  -[-  aq  =  As  0, -\-  aq 

+  2  W  O4. 
The  symbol  of  crystallized  binarseniate  of  potassa 
is  [2  H  O,  K  O]  As  O5. 

Yellow. 

416.  As  03-\-d  H  S-\-Aq  =  As^s+M- 

Yellow. 

As  0,-\-?>  HS^  Aq=  A&^,-\-  Aq. 

Yellow.  Red. 

2  As  S3  +  S  =  2  As  Sj. 
10* 


114  HEAVY  METALS. 

416.  The   symbol  of  Mispickel   (arsenical    pyrites)   is 

Fe  [As,   SJ. 
2  (Fe  [As,  S^])  +  13  O  =  Fea  O3  +  2  As  O3 
+  2SO2. 

417.  As  Zng  +  3  {HO,  SO,)-\-Aq^Z  {Zn  0,  S  0,) 

+  ^9'  +  As  H3. 

418.  Sb  Zng  +  3  (^0,  aS  O3)  +  ^^  =  3  {Zk  0,  S  O3) 

^Aq-\-Sh  Ha. 
The  compounds  of  arsenic  and  antimony  with  hy- 
drogen are  always  mixed  with  more  or  less  free 
hydrogen.  When  prepared  as  described  in 
sections  417,  418  of  Stockhardt's  Elements,  the 
gas  consists  almost  entirely  of  hydrogen,  con- 
taining only  a  very  minute  amount  of  either  me- 
tallic compound. 


TABLES. 


EXPLANATION    OF    TABLES. 


Table  I.  —  This  table,  Tvhich  has  been  reprinted  from  the  "  Elementary 
Instructions  in  Chemical  Analysis"  by  Fresenius,  indicates  by  means  of 
figures  the  solubility  or  insolubility  in  water  and  acids  of  some  of  the  more  fre- 
quently occurring  compounds;  thus,  1  means  a  substance  soluble  in  water; 
2,  a  substance  insoluble  in  water,  but  soluble  in  chlorohydric  or  nitric  acid; 
8,  a  substance  insoluble  either  in  water  or  acids.  For  those  substances  stand- 
ing on  the  limits  between  these  three  classes,  the  figures  are  jointly  expressed; 
thus  1-2  signifies  a  substance  difficultly  soluble  in  water,  but  soluble  in 
chlorohydric  or  nitric  acid ;  1  -  3,  a  body  difficultly  soluble  in  water,  and  the 
solubility  of  which  is  not  increased  on  the  addition  of  acids ;  and  2  -  3,  a  sub- 
stance insoluble  in  water,  and  difficultly  soluble  in  the  acids.  When  the  rela- 
tion of  a  substance  to  hydrochloric  acid  is  different  from  that  to  nitric  acid, 
this  is  stated  in  the  notes.  The  figure  indicating  the  solubility  of  a  given  salt 
will  be  found  opposite  to  the  symbol  of  its  acid,  in  the  column  headed  by  the 
symbol  of  its  base;  that  of  a  given  binary,  under  the  symbol  of  the  correspond- 
ing oxide,  and  opposite  to  the  symbol  of  its  electro-negative  element. 

Table  II.  —  The  values  of  the  French  measures  and  weights,  in  terms  of  the 
corresponding  English  units,  given  in  this  table,  were  taken  from  the  second 
volume  of  the  Cavendish  Edition  of  "  Gmelin's  Hand-Book  of  Chemistry." 
The  logarithms  of  these  values  and  their  arithmetical  complements  have  been 
added  to  facilitate  the  reduction  from  one  system  to  the  other.  The  use  of  the 
table  can  be  illustrated  best  by  a  few  examples. 

1.  It  is  required  to  reduce  560.367  metres  to  English  feet. 
Solution.  —  No.  of  feet  =  No.  of  metres  X  No.  of  feet  in  one  metre. 

log.  No.  of  feet  =:  log.  No.  of  metres  +  log.  No.  of  feet  in  one  metre, 
log.  560.367  2.7484726 

log.  3.2809  (value  in  feet  of  one  metre  from  Table  II.)      0.5159930 

"  3.2644656 

Ans.  =  1838.51  feet. 


118  EXPLANATION   OF   TABLES. 

2.  It  is  required  to  reduce  30.964  inches  to  centimetres. 

Solution.  —  No.  of  centimetres  =  No.  of  inches  -^  No.  of  inches  in  one  cen- 
timetre, 
log.  No.  of  centimetres  =  log.  No.  of  inches  +  log.  (ar.  co.)  No.  of 

inches  in  one  centimetre, 
log.  30.964  1.4908571 

log.  (ar.  CO.)  0.3937  (value  of  one  centimetre  in  inches)    0.4048^58 

1.8956829 
Ans.  78.6471  centimetres. 

3.  It  is  required  to  reduce  23.576  kilometres  to  feet. 
Solution.  —  23.576  kilometres  =  23576  metres. 

No.  of  feet  =  No.  of  metres  X  No.  of  feet  in  one  metre.  ' 

log.  2.3576  4.3724701. 

log.  3.2809  0.5159930 

4.8884631 
Ans.  77350.5  feet. 

In  the  above  examples  logarithms  of  seven  places  have  been  used ;  but  where 
great  accuracy  is  not  required,  logarithms  of  four  places  are  sufficient.  In 
such  cases  the  last  three  figures  of  the  logarithm  given  in  the  table  may  be 
neglected,  and  the  problems  solved  with  great  expedition  by  means  of  the  table 
of  four-place  logarithms  which  accompanies  this  book. 

Table  III.  —  This  table  has  been  taken,  with  some  few  alterations,  from 
Weber's  "  Atomgewichts-Tabellen."  The  atomic  volumes  assigned  to  the  ele- 
ments are  the  same  as  those  generally  given  in  English  and  American  text- 
books on  Chemistrj',  with  the  exception  of  those  of  Carbon,  Boron,  and  Silicon, 
which  are  assumed  to  yield  a  one-volume  gas  like  oxygen  for  convenience  in 
calculation.  The  calculated  specific  gravities  are  deduced  from  the  observed 
specific  gravity  of  ox3'gen  and  the  chemical  equivalent  of  the  given  substance 
by  means  of  the  proportion, 

Equiv.  of  Oxygen  :  Equiv.  of  given  substance  =  Sp.  Gr.  of  Oxygen  :  Sp.  Gr. 
of  given  substance. 

This  proportion  yields  the  specific  gravity  directly  when  one  equivalent  of 
the  substance  occupies  the  same  volume  as  one  equivalent  of  oxygen.  If  it 
occupies  twice,  three  times,  or  four  times  this  volume,  the  results  must  be 
divided  by  two,  three,  or  four,  as  the  case  may  be.  The  method  of  calculating 
may  best  be  illustrated  by  a  few  examples. 

1.  It  is  required  to  calculate  the  Specific  Gravity  of  Nitrogen.  ' 

Equiv.  of  0.  Equiv.  ofN.  Sp.  Gr.  of  O. 

Solution.  8        :        14      =      1.10563     :     1.93485. 

This  would  be  the  specific  gravity  if  14  parts  of  nitrogen  occupied  the  same 
volume  as  8  parts  of  oxygen ;  or,  in  other  words,  if  the  equivalent  volume  of 


EXPLANATION    OF   TABLES.  119 

nitrogen  was  1,  the  same  as  that  of  oxygen.    The  fact  is  that  it  is  2,  so  that 
the  true  specific  gravity  of  nitrogen  =  t  (1.93485)  =  0.967428. 

2.  It  is  required  to  calculate  the  specific  gravity  of  ammonia  gas. 

EquiT.  ofO.    Equiv.  ofNHj.  Sp.  Gr.  ofO. 

Solution.  8        :         17        =         1.10563     :     2.34946. 

Hence  the  specific  gravity  of  ammonia  gas  •would  be  2.34946  if  one  equiva- 
lent (or  17  parts)  occupied  the  same  volume  as  one  equivalent  of  oxygen;  but 
on  referring  to  the  table,  it  will  be  found  that  the  equivalent  volume  of  this  gas 
is  4,  or,  in  other  words,  that  one  equivalent  occupies  a  volume  four  times  as 
large  as  that  occupied  by  one  equivalent  of  oxygen,  so  that  to  find  the  true 
specific  gravity  of  ammonia  gas  we  must  divide  2.34946  by  4,  which  will  give 
a  quotient  0.58736.  The  slight  difference  between  this  result  and  that  given 
in  the  table  arises  from  the  fact  that  in  Weber's  table  the  equivalent  of  nitro- 
gen used  is  14.005,  and  not  14,  as  in  the  solutions  of  the  above  examples. 

Were  the  law  of  equivalent  volumes  absolutely  rigorous,  (that  is,  did  one 
equivalent  of  every  gas  precisely  occupy  either  the  same  volume,  or  else  a  vol- 
ume two,  three,  or  four  times  as  great  as  the  volume  of  oxygen,)  then  the  cal- 
culated specific  gravities  ought  to  agree  exactly  with  those  obtained  by  experi- 
ment. On  comparing  together  the  two  columns  of  observed  and  calculated 
specific  gravities  in  the  table,  it  will  be  found  that  the  numbers,  although 
approximatively  equal,  do  not  absolutely  coincide.  Part  of  these  differences 
are  unquestionably  owing  to  errors  of  observation ;  but  after  making  the  great- 
est possible  allowance  for  all  errors  of  that  sort,  there  still  remains  (especially 
in  the  case  of  the  easily  condensed  gases,  such  as  alcohol  vapor,  sulphurous 
acid,  and  carbonic  acid)  large  differences  to  be  accounted  for.  The  most  prob- 
able explanation  of  these  differences  seems  to  be  found  in  the  assumption  that 
the  law  of  equivalent  volumes  holds  rigorously  only  when  the  gases  are  in  the 
state  of  extreme  expansion.  As  we  experiment  upon  them,  they  are  more  or 
less  condensed  by  the  pressure  of  the  atmosphere,  and  it  is  supposed  that  they 
are  not  all  condensed  equally,  or,  in  other  words,  that  even  under  this  pressure 
they  do  not  obey  absolutely  the  law  of  Marriotte.  The  more  easily  a  gas  may 
be  reduced  to  a  fluid,  the  greater  is  it  condensed  by  the  atmospheric  pressure, 
and  hence  the  greater  is  its  specific  gravity.  This  view  is  confirmed  by  the 
fact  that  the  observed  specific  gravity  of  carbonic  acid  gas  at  0°,  and  under  a 
more  feeble  pressure  than  that  of  the  atmosphere,  approaches  more  nearly  to 
that  obtained  by  experiment. 

The  Specific  gravity  of  Carbonic  Acid  Gas,  at  0°  (air  =  1),  was. 

Under  the  pressure  of  76.000  centimetres,  1.52910 

"         "         "         37.413         "  1.52366 

"  "  "  22.417*      "  1.52145 

Theoretical  specific  gravity,  1.52024 

*  It  must  be  remembered  that  the  specific  gravity  of  a  gas  is  equal  to  its  weight,  di- 
vided by  the  weight  of  an  equal  volume  of  air  under  the  same  conditions  of  temperature 
and  pressure. 


120 


TABLES. 


^       I 

O 
I— I 

iJ 
O 


o 

i-Ih 

CO                G<J                                                           (N    (N 
(M     |<MCO     |<>?THffq<N(>»(MTHG<l       |        | 

CN                 1—1                                                               tH     I— 1 

o 

l-H 

o 

s 

O 

1—1                     t-l 

o 

(MrHrH(MT-(          r-l(MCCi          G<><J^G<JfH(M 

O 

<Mi-(r-l(Mi-lrHT-l(M(M          (M(N(NtH       | 

o 

CO                       CO 
r-l               •         CM 

O 

tHrHT-lTHC0<?^TH<M<N<N5<l<?^(n     rH     <M 

o 

in 

tHrHTHrHCO(Mi-l<MCNi-H(NCN(M     i-l    (M 

o 

(M                (M  CO 

T-l                    iH   1— 1 

o 

be 

<1 

jcoco<N    |g^ih<m(N(»(M(Mit^   ih   cm 

r-t                        r-l 

o 
M  ■ 

o 

o 

1 — 1 
1—1 

OhHC/2ccO^ptH<|<lOpq0|<1lH 

121 


(M     T-i 

(MrH<>?iH          (M         (N(?^C<Ir-((M 

++ 

<M  CN  C<1                 (Ja  <M                    1      i-H     i-H 

6 

tH 

O 

5<l  i-H 

CO   l-H            t-^ 

6 

CO 

++ 

O 

|>1  tH 

<>?  tH  <M                                    (M  <3<1     iH       1 

6 

CO 

1— lr-liHG<l(M                 (m|i— liH 
tH 

O 

J? 

CO 

!H(NtH<Mi-l                  G<Jr-lrHTH 

c5 

CO 

iHCNrHCMfKI                  «<l(?qr-<r-l 

o 

(M  j-l 

-t— 

o 

(N  1—1 

o 

tH               r-l 

o 

1 

<Mt-(THS<JtHtHiH(M(M         (I<1(M<M                | 

Om 

is 

n 

>-, 

,n 

:S 

-7^ 

y. 

o 

S 

03 

pi 

t>i 

rO 

O 

ro 

.9 

^ 

^ 

f  5 

O. 

a 

d 

rS 

a> 

P 

43 

'« 

!> 

[>,H 

rC 

o 

03 

d 

"i^ 

O 

^ 

CJ 

^rO 

a 

^ 

n 

'S 

^ 

g 

o 

■73 

•  l-H 

1^ 

>% 

r" 

^ 

o 

"a-Q 

C) 

n 

o 

n-f 

H 

s 

s 

t> 

4^ 

rQrQ 

13 

o 

^i 

'C  itJ" 

o 

-H 

•ri 

•p! 

^ 

'« 

— 1 

~ 

(=-.o 

o 

<■> 

a 

n1 

.£3 

s 

*r^ 

h 

'S 

'S 

13 

d 
o 

d 

fl 

o 

d 

r2 

<D 

o 

o 

0> 

^-^ 

o 

n 

-> 

ccco 

Q!z;c» 

11 


122 


TABLES. 


TABLE    II. 


FKENCH  MEASUEES  AND  WEIGHTS. 
Measures  of  Length. 


1  Kilometre 

1  Hectometre 

1  Decimetre 

1  Metre 


1  Kilometre 

1  Metre 

1  Centimetre 


1000  Metres. 

100       « 

10       " 

1       " 

0.6214  Miles. 

3.2809  Feet. 

0.3937  Inches 

1  Meti-e  =  1.000  Metre. 

1  Decimetre   =  0.100      " 

1  Centimetre  =  0.010      " 

1  MilUmetre   =  0.001      " 


Logarithms. 
9.7933712 
0.5159930 
9.5951742 


Ar.  Co.  Log. 
0.2066188 
9.4840070 
0.4048258 


Measures  of  Volume. 

1  Cubic  Metre  =     1000.000  Litres. 

1  Cubic  Decimetre     =  1.000       " 

1  Cubic  Centimetre  =  0.001       " 


Logarithms. 

Ar.  Co.  Log, 

1  Cubic  Metre 

=  35.31660  Cubic  Feet. 

1.5479790 

8.4520210 

1  Cubic  Decimetre 

=  61.02709  Cubic  Inches. 

1.7855226 

8.2144774 

1  Cubic  Centimetre 

=     0.06103        " 

8.7855226 

1.2144774 

1  Litre 

=     0.22017  Gallons. 

9,3427581 

0.6572419 

1  Litre 

=     0.88066  Quarts. 

9.9448083 

0.0551917 

1  Litre 

=     1.76133  Pints. 
Weights. 

0.2458407 

9.7541593 

1  Kilogramme    =  1000  Grammes. 

1  Hectogramme  =    100         " 

1  Decagramme   =10         " 

1  Gramme  =        1  " 


1  Gramme         =  1,000  Gramme. 

1  Decigramme   =  0.100        " 

1  Centigramme  =  0.010         " 

1  Milligramme  =  0.001        " 


1  Kilogramme    =     2.67951  Pounds  (Troy). 
1  Gramme  =   15.44242  Grains. 


Logarithms.  Ar,  Co.  Log, 
0.4280554  9.5719446 
1.1887154         8.8112846 


TABLES. 


123 


TABLE    III. 

SPECIFIC  GRAVITY  AND  ABSOLUTE  WEIGHT  OF  ONE 
LITRE  OF  SOME  OF  THE  MOST  IMPORTANT  GASES 
AND   VAPORS. 


Names  of  Gases. 

> 
'3 

Sp.  Gr. 
Observed. 

Sp.  Gr. 
Calculated. 

Weight  of 
1  Litre  = 
1000  c.  c. 

Loga- 
rithms. 

Ar.  Co.  Log- 
arithms. 

Air, 

1. 

1.00000 

1.29363 

0.1118101 

9.8881899 

Alcohol, 

4 

1.613 

1.58934 

2.05602 

0,3130273 

9.6869727 

Ammonia  Gas, 

4 

0.5967 

0.58753 

0.76005 

9.8808422 

0.1191578 

Antimony, 

1 

17.83274 

23.06897 

1.3630282 

8.6369718 

Antimonide  of  Hydr. 

4 

4.56239 

5.90204 

0.7710022 

9.2289978 

Arsenic, 

1 

10.65 

10.36528 

13.40884 

1.1273912 

8.8726088 

Arsenide  of  Hydr. 

4 

2.695 

2.69553 

3.48702 

0.5424519 

9.4575481 

Boron, 

1 

1.50591 

1.94809 

0.2896090 

9.7103910 

Bromine, 

2 

5.54 

5.52605 

7.14866 

0.8542247 

9.1457753 

Bromoliydric  Acid, 

4 

2.79758 

3.61903 

0.5585922 

9.4414078 

Carbon, 

1 

0.8469* 

0.82922 

1.07270 

0.0304783 

9.9695217 

Carbonic  Acid, 

2 

1.52908 

1.52024 

1.96663 

0.2937226 

9.7062774 

Carbonic  Oxide, 

2 

0.96779 

0.96743 

1.25150 

0.0974309 

9.9025691 

Chlorine, 

2 

2.47 

2.45052 

3.17007 

0.5010689 

9.4989311 

Chloride  of  Boron, 

4 

3.942 

4.05226 

5.24213 

0.7195078 

9.2804922 

Chloride  of  Silicon, 

3 

5.939 

5.92477 

7.66446 

0.8844816 

9.1155184 

Chlorohydric  Acid, 

4 

1.2474 

1.25981 

1.62973 

0.2121157 

9.7878843 

Cyanogen, 

2 

1.8064 

1.79698 

2.32463 

0.3663538 

9.6336462 

Cyanhydric  Acid, 

4 

0.9476 

0.93304 

1.20701 

0.0817109 

9.9182891 

Ether, 

2 

2.586 

2.55677 

3.30751 

0.5195012 

9.4804988 

Fluorine, 

2 

1.30151 

1.68367 

0.2262570 

9.7737430 

Fluoride  of  Boron, 

4 

2.3124 

2.32875 

3.01254 

0.4789329 

9.5210671 

Fluoride  of  Silicon, 

3 

3.600 

3.62677 

4.69170 

0.6713302 

9.3286698 

Fluohydric  Acid, 

4 

0.68531 

0.88654 

9.9476983 

0.0523017 

Hydrogen, 

2 

0,06927 

0.06910 

0.08939 

8.9512889 

1.0487111 

Iodine, 

2 

8.716 

8.76760 

11.34203 

1.0547011 

8.9452989 

lodohydric  Acid, 

4 

4.443 

4.41835 

5.71571 

0.7570702 

9.2429298 

Marsh  Gas, 

4 

0.5576 

0.55282 

0.71514 

9.8543911 

0.1456089 

Mercuiy, 

2 

6.976 

6.91732 

8.94845 

0.9517478 

9.0482522 

Nitrogen, 

2 

0.97136 

0.96776 

1.25192 

0.0975765 

9.9024235 

Nitrous  Oxide, 

2 

1.5269 

1.58951 

2.05624 

0.3130738 

9.6869262 

Nitric  Oxide, 

4 

1.0388 

1.03669 

1.34109 

0.1274580 

9.8725420 

defiant  Gas, 

4 

0.9852 

0.96743 

1.25150 

0.0974309 

9.9025691 

Oxygen, 

1 

l.li)§63 

1.43028 

0.1554210 

9.8445790 

Phosphoms, 

1 

4.42 

4.33452 

5.60727 

0.7487515 

9.2512485 

Phosphide  of  Hydr. 

4 

1.178 

1.18728 

1.53590 

0.1863629 

9.8136371 

Selenium, 

1 

2.73801 

3.54197 

0.5492448 

9.4507552 

Silicon, 

1 

3.07120 

3.97300 

0.5991186 

9.4008814 

Sulphur, 

k 

6.5635 

6.65866 

8.61384 

0.9351968 

9.064S032 

Sulphide  of  Hydr. 

2 

1.1912 

1.17888 

1.52.503 

0.1832784 

9.8167216 

Sulphurous  Acid, 

2 

2.247 

2.21541 

2.86592 

0.4572640 

9.5427360 

*  Calculated  from  the  Sp.  Gr.  of  Carbouic  Acid  Gas,  observed  by  RegnauU. 


124 


TABLES. 


TABLE    IV. 

MEAN  COEFFICIENTS  OF  LINEAE  EXPANSION  OF  SOLIDS 
FOR   ONE  DEGEEE  BETWEEN  0°  AND   100°. 


Name  of  Substance. 

Coefficients. 

Name  of  Observer. 

Glass  (flint  of  Clioisy  le  Roi), 

0.00000760 

Regnault. 

Platinum, 

0.00000884 

Dulong  and  Petit. 

Glass  (commoa  of  Paris), 

0.00000920 

Regnault. 

Palladium, 

0.00001000 

WollastOn. 

Antimony, 

0.00001083 

Smeaton. 

Iron  (soft  forged), 

0.00001220 

Lavoisier  and  Laplace. 

Bismuth, 

0.00001392 

Smeaton. 

Gold, 

0.00001466 

Lavoisier  and  Laplace. 

Brass  (English,  in  rods), 

0.00001893 

Roy. 

Copper, 

0.00001919 

Troughton. 

Silver, 

0.00002083 

« 

Tin  (fine), 

0.00002283 

<i 

Lead, 

0.00002866 

a 

Zinc, 

0.00002942 

a 

APPARENT   CUBIC   EXPANSION  OF  LIQUIDS  IN  GLASS 
BETWEEN  0°  AND   100°. 


Name  of  Substance. 

Expansion  from  Oo  to  lOOo. 

Water, 

A 

— 

0.0466 

Chlorohydric  Acid,  Sp.  Gr.  1.137, 

2V 

= 

0.0600 

Nitric  Acid,  Sp.  Gr.  1.40, 

* 

= 

0.0100 

Sulphuric  Acid,  Sp.  Gr.  1.85, 

tV 

= 

0.0600 

Common  Ether, 

-h 

= 

0.0700 

Olive  Oil, 

-h 

= 

0.0800 

Oil  of  Tui-pentine, 

-h 

= 

0.0700 

Water  saturated  with  salt, 

■h 

= 

0.0500 

Alcohol, 

1 
9 

= 

0.1100 

Mercury, 

-h 

= 

0.0156 

COEFFICIENTS   OF  CUBIC  EXPANSION  OF  GASES. 
Observed  bt  V.    REGNAULT. 


Name  of  Substance. 

Under  constant  Volume. 

Under  constant  Pressure. 

Air, 

0.003665 

0.003670 

Nitrogen, 

0.003668 

0.003670 

Hydrogen, 

0.003667 

0.003661 

Oxide  of  Carbon, 

0.003667 

0.003669 

Carbonic  Acid, 

0.003688 

0.003710 

Cyanogen, 

0.003829 

0.003877 

Sulphm-ous  Acid, 

0.003845 

0.003903 

TABLES. 


125 


TABLE    V. 

DENSITIES  AND  VOLUMES  OF  WATER. 

By  M.  DESPRETZ. 


as 

S3 

Volumes. 

Densities. 

Volumes. 

Densities. 

Volumes. 

Densities. 

0 

1.0001269 

0.999873 

34 

1.00555 

0.994480 

68 

1.02144 

0.979010 

1 

1.0000730 

0.999927 

35 

1.00593 

0.994104 

69 

1.02200 

0.978473 

2 

1.0000331 

0.9999661 

36 

1.00624 

0.993799 

70 

1.02255 

0.977947 

3 

1.0000083 

0.999999 

37 

1.00661 

0.993433 

71 

1.02315 

0.977373 

4 

1.0000000 

1.000000 

38 

1.00699 

0.993058 

72 

1.02375 

0.976800 

5 

1.0000082 

0.999999 

39 

1.00734 

0.992713 

73 

1.02440 

0.976181 

6 

1.0000309 

0.999969 

40 

1.00773 

0.992329 

74 

1.02499 

0.975619 

7 

1.0000708 

0.999929 

41 

1.00812 

0.991945 

75 

1.02562 

0.975018 

8 

1.0001216 

0.999878 

42 

1.00853 

0.991542 

76 

1.02631 

0.974364 

9 

1.0001879 

0.999812 

43 

1.00894 

0.991139 

77 

1.02694 

0.973766 

10 

1.0002684 

0.999731 

44 

1.00938 

0.990707 

78 

1.02761 

0.973132 

11 

1.0003598 

0.999640 

45 

1.00985 

0.990246 

79 

1.02823 

0.972545 

12 

1.0004724 

0.999527 

46 

1.01020 

0.989903 

80 

1.02885 

0.971959 

13 

1.0005862 

0.999414 

47 

1.01067 

0.989442 

81 

1.02954 

0.971307 

14 

1.0007146 

0.999285 

48 

1.01109 

0.989032 

82 

1.03022 

0.970666 

15 

1.0008751 

0.999125 

49 

1.01157 

0.988562 

83 

1.03090 

0.970027 

16 

1.0010215 

0.998979 

50 

1.01205 

0.988093 

84 

1.03156 

0.969405 

17 

1.0012067 

0.998794 

51 

1.01248 

0.987674 

85 

1.03225 

0.968757 

18 

1.00139 

0.998612 

52 

1.01297 

0.987196 

86 

1.03293 

0.968120 

19 

1.00158 

0.998422 

53 

1.01345 

0.986728 

87 

1.03351 

0.967482 

20 

1.00179 

0.998213 

154 

1.01395 

0.986243 

88 

1.03430 

0.966837 

21 

1.00200 

0.998004 

55 

1.01445 

0.985756 

1  89 

1.03500 

0.966183 

22 

1.00222 

0.997784 

56 

1.01495 

0.985270 

90 

1.03566 

0.965567 

23 

1.00244 

0.997566 

57 

1.01547 

0.984766 

91 

1.03639 

0.964887 

24 

1.00271 

0.997297 

58 

1.01597 

0.984281 

92 

1.03710 

0.964227 

25 

1.00293 

0.997078 

59 

1.01647 

0.983798 

93 

1-03782 

0.963558 

26 

1.00321 

0.996800 

60 

1.01698 

0.983303 

94 

1.03852 

0.962908 

27 

1.00345 

0.996562 

61 

1.01752 

0.982782 

95 

1.03925 

0.962232 

28 

1.00374 

0.996274 

62 

1.01809 

0.982231 

96 

1.03999 

0.961547 

29 

1.00403 

0.995986 

63 

1.01862 

0.981720 

97 

1.04077 

0.960827 

30 

1.00433 

0.995688 

64 

1.01913 

0.981229 

j  98 

1.04153 

0.960125 

31 

1.00463 

0.995391 

65 

1.01967 

0.980709 

99 

1.04228 

0.959434 

32 

1.00494 

0.995084 

66 

1.02025 

0.980152 

jlOO 

1.04315 

0.958634 

33  1  1.00525 

0.994777 

67 

1.02085 

0.979576 

126 


TABLES. 


TABLE    VI. 

PER   CENT   OF  N  O5    IN  AQUEOUS    SOLUTIONS   OF  DIF- 
FERENT   SPECIFIC   GRAVITIES.    By  URE.     15°  C. 


Specific 

Per  Cent 

Specific 

Per  Cent 

Specific 

Per  Cent 

Specific 

Per  Cent 

Gravity. 

of  N  O5. 

Gravity. 

ofNOg. 

Gravity. 

of  N  Og. 

Gravity. 

ofNOj. 

1.500 

79.7 

1.419 

59.8 

1.295 

39.8 

1.140 

19.9 

1.498 

78.9 

1.415 

59.0 

1.289 

39.0 

1.134 

19.1 

1.496 

78.1 

1.411 

58.2 

1.283 

38.3 

1.129 

18.3 

1.494 

77.3 

1.406 

57.4 

1.276 

37.5 

1.123 

17.5 

1.491 

76.5 

1.402 

56.6 

1.270 

36.7 

1.117 

16.7 

1.488 

75.7 

1.398 

55.8 

1.264 

35.9 

1.111 

15.9 

1.485 

74.9 

1.394 

55.0 

1.258 

35.1 

1.105 

1.5.1 

1.482 

74.1 

1.388 

54.2 

1.252 

34.3 

1.099 

14.3 

1.479 

73.3 

1.383 

53.4 

1.246 

33.5 

1.093 

13.5 

1.476 

72.5 

1.378 

52.6 

1.240 

32.7 

1.088 

12.7 

1.473 

71.7 

1.373 

51.8 

1.234 

31.9 

1.082 

11.9 

1.470 

70.9 

1.368 

51.1 

1.228 

31.1 

1.076 

11.2 

1.467 

70.1 

1.363 

50.2 

1.221 

30.3 

1.071 

10.4 

1.464 

69.3 

1.358 

49.4 

1.215 

29.5 

1.065 

9.6 

1.460 

68.5 

1.3.53 

48.6 

1.208 

28.7 

1.059 

8.8 

1.457 

67.7 

1.348 

47.9 

1.202 

27.9 

1.054 

8.0 

1.453 

66.9 

1.343 

47.0 

1.196 

27.1 

1.048 

7.2 

1.450 

66.1 

1.338 

46.2 

1.189 

26.3 

1.043 

6.4 

1.446 

65.3 

1.332 

45.4 

1.183 

25.5 

1.037 

5.6 

1.442 

64.5 

1.327 

44.6 

1.177 

24.7 

1.032 

4.8 

1.439 

63.8 

1.322 

43.8 

1.171 

23.9 

1.027 

4.0 

1.435 

63.0 

1.316 

43.0 

1.165 

23.1 

1.021 

3.2 

1.431 

62.2 

1.311 

42.2 

1.159 

22.3 

1.016 

2.4 

1.427 

61.4 

1.306 

41.4 

1.153 

21.5 

1.011 

1.6 

1.423 

60.6 

1.300 

40.4 

1.146 

20.7 

1.005 

0.8 

PER   CENT   OF  H  CI  IN    AQUEOUS    SOLUTIONS    OF  DIF- 
FERENT  SPECIFIC  GRAVITIES.    By  E  DAVY     15°  C. 


Specific  Gravity. 

Per  Cent  of  H  CI. 

1 

i     SiJecific  Gravity, 

Per  Cent  of  H  CI. 

1.21 

42.43 

110 

20.20 

1.20 

40.80 

1.09 

18.18 

1.19 

38.38 

1.08 

16.16 

1.18 

36.36 

1.07 

14.14 

1.17 

34.34 

1.06 

12.12 

1.16 

32.32 

1.05 

10.10 

1.15 

30.30 

1.04 

8.08 

1.14 

28.28 

1.03 

6.06 

1.13 

26.26 

1.02 

4.04 

1.12 

24.24 

1.01 

2.02 

1.11 

22.22 

TABLES. 


127 


PER  CENT  OF  HO,  S  O3  AND  S  O3  IN  AQUEOUS  SO- 
LUTIONS or  DIEEEEENT  SPECIFIC  GEAVITIES.  By 
BINEAU.    15°  C. 


Per  Cent  of 

Specific 

Per  Cent  of 

Percent  of 

Specific 

Per  Cent  of 

HO,  SO3. 

Gravity. 

SO3. 

H  0,  S  O3 

Gravity. 

SO3. 

100 

1.8426 

81.63 

56 

1.4586 

45.71 

99 

1.842 

80.81 

55 

1.448 

44.89 

98 

1.8406 

80.00 

54 

1.438 

44.07 

97 

1.840 

79.18 

53 

1.428 

43.26 

96 

1.8384 

78.36 

52 

1.418 

42.45 

95 

1.8376 

77.55 

51 

1.408 

41.63 

94 

1.8356 

76.73 

50 

1.398 

40.81 

93 

1.834 

75.91 

49 

1.3886 

40.00 

92 

1.831 

75.10 

48 

1.379 

39.18 

91 

1.827 

74.28 

47 

1.370 

38.36 

90 

1.822 

73.47 

46 

1.361 

37.55 

89 

1.816 

72.65 

45 

1.351 

36.73 

88 

1.809 

71.83 

44 

1.342 

35.82 

87 

1.802 

71.02 

43 

1.333 

35.10 

86 

1.794 

70.10 

42 

1.324 

34.28 

85 

1.786 

69.38 

41 

1.315 

33.47 

i         84 

1.777 

68.57 

40 

1.306 

32.65 

!         83 

1.767 

67.75 

39 

1.2976 

31.83 

82 

1.756 

66.94 

38 

1.289 

31.02 

81 

1.745 

66.12 

37 

1.281 

30.20 

80 

1.734 

65.30 

36 

1.272 

29.38 

79 

1.722 

64.48 

35 

1.264 

28.57 

78 

1.710 

63.67 

34 

1.256 

27.75 

77 

1.698 

62.85 

33 

1.2476 

26.94 

76 

1.686 

62.04 

32 

1.239 

26.12 

75 

1.675 

61.22 

31 

1.231 

25.30 

74 

1.663 

60.40 

30 

1.223 

25.49 

73 

1.651 

59.59 

29 

1.215 

23.67 

72 

1.639 

58.77 

28 

1.2066 

22.85 

71 

1.637 

57.95 

27 

1.198 

22.03 

70 

1.615 

57.14 

26 

1.190 

21.22 

69 

1.604 

56.32 

25 

1.182 

20.40 

68 

1.592 

55.59 

24 

1.174 

19.58 

67 

1.580 

54.69 

22 

1.159 

17.95 

66 

1.578 

53.87 

20 

1.144 

16.32 

65 

1.557 

53.05 

18 

1.129 

14.69 

64 

1.545 

52.24 

16 

1.1136 

13.06 

63 

1.534 

51.42 

14 

1.098 

11.42 

62 

1.523 

50.61 

12 

1.083 

9.79 

61 

1.512 

49.79 

10 

1.068 

8.16 

60 

1.501 

48.98 

8 

1.0536 

6.53 

59 

1.490 

48.16 

6 

1.039 

4.89 

58 

1.480 

47.34 

4 

1.0256 

3.26 

57 

1469 

46.53 

2 

1.013 

1.63 

128 


TABLES. 


PER  CENT  OF  NH3  IN  AQUEOUS   SOLUTIONS  OF  DIF- 
FERENT SPECIFIC  GRAVITIES.    By  J.  OTTO.     16°  C 


Specific 

Per  Cent  of 

Specific 

Per  Cent  of 

Specific 

Per  Cent  of 

Gravity. 

NH3. 

Gravity. 

NH3. 

Gravity. 

NH3. 

0.9517 

12.000 

0.9607 

9.625 

0.9697 

7.250 

0.9521 

11.875 

0.9612 

9.500 

0.9702 

7.125 

0.9526 

11.750 

0.9616 

9.375 

0.9707 

7.000 

0.9531 

11.625 

0.9621 

9.250 

0.9711. 

6.875 

0.9536 

11.500 

0.9626 

9.125 

0.9716 

6.750 

0.9540 

11.375 

0.9631 

9.000 

0.9721 

6.625 

0.9545 

11.250 

0.9636 

8.875 

0.9726 

6.500 

0.9550 

11.125 

0.9641 

8.750 

0.9730 

6.375 

0.9555 

11.000 

0.9645 

8.625 

0.9735 

6.250 

0.9556 

10.950 

0.9650 

8.500 

0.9740 

6.125 

0.9559 

10.875 

0.9654 

8.375 

0.9745 

6.000 

0.9564 

10.750 

0.9659 

8.250 

0.9749 

5.875 

0.9569 

10.625 

0.9664 

8.125 

0.9754 

5.750 

0.9574 

10.500 

0.9669 

8.000 

0.9759 

5.625 

0.9578 

10.375 

0.9673 

7.875 

0.9764 

5.500 

0.9583 

10.250 

0.9678 

7.750 

0.9768 

5.375 

0.9588 

10.125 

0.9683 

7.625 

0.9773 

5.250 

0.9593 

10.000 

0.9688 

7.500 

0.9778 

5.125 

0.9597 

9.875 

0.9692 

7.375 

0.9783 

5.000 

0.9602 

9.750 

TABLE    VII. 

PER  CENT  OF  OXYGEN  IN  DIFFERENT  OXIDES. 


Oxides. 

Per  Cent. 

Logarithms. 

Oxides. 

Per  Cent. 

Logaritlims. 

AI2O3 

0.467 

9.6684 

KO 

0.170 

9.2304 

AsOs 

0.348 

9.5416 

LiO 

0.550 

9.7404 

BO3 

0.688 

9.8376 

MO3 

0.343 

9.5353 

BaO 

0.104 

9.0170 

MgO 

0.400 

9.6021 

BeaOs 

0.630 

9.7993 

MnO 

0.225 

9.3522 

BiaOs 

0.103 

9.0128 

Mna  O3 

0.303 

9.4814 

COa 

0.727 

9.8615 

NO5 

0.740 

9.8692 

Ca  0 

0.286 

9.4564 

NaO 

0.258 

9.4116 

CoO 

0.213 

9.3284 

NiO 

0.213 

9.3284 

Crs  O3 

0.473 

9.6749 

PO5 

0.563 

9.7505 

CuaO 

0.112 

9.0492 

PbO 

0.717 

9.8555 

CuO 

0.201 

9.3032 

SiOs 

0.530 

9.7243 

FeO 

0.222 

9.3464 

SrO 

0.154 

9.1875 

FeaOs 

0.300 

9.4771 

TaOs 

0.115 

9.0607 

HO 

0.889 

9.9489 

ZnO 

0.197 

9.2945 

ANTILOGARITHMS. 

.  a 

Proportional  Parts.   | 

n 

0 

1 

2 

3 

4 

5 

6 

7 

8 

9 

II 

1019 

1  2 

0  0 

s 

4 

5 

6 

1 

7 
2 

8 

2 

9 
2 

.00 

1000 

1002 

1005 

1007 

1009 

1012 

1014 

1016 

1021 

.01 

1023 

1026  1028:1030 

1033 

1035 

1038  1040 

1042 

1045 

0  0 

1 

2 

2 

2 

.02 

1047 

1050 

1052  1054 

1057 

1059 

1062 

1064 

1067 

1069 

0  0 

1 

2 

2 

2 

.03 

1072 

1074 

1076 

1079 

1081 

1084 

1086 

1089 

1091 

1094 

0  0 

1 

2 

2 

2 

.04 

1096 

1099 

1102 

1104 

1107 

1109 

1112 

1114 

1117 

1119 

0  1 

2 

2 

2 

2 

.05 

1122 

1125 

1127 

1130 

1132 

1135 

1138 

1140 

1143 

1146 

0  1 

2 

2 

2 

2 

.06 

1148 

1151 

1153 

1156 

1159 

1161 

1164 

1167 

1169 

1172 

0  1 

2 

2 

2 

2 

.07 

1175 

1178 

1180 

1183 

1186 

1189 

1191 

1194 

1197 

1199 

0  1 

2 

2 

2 

2 

.08 

1202 

1205 

1208 

1211 

1213 

1216 

1219 

1222 

1225 

1227 

0  1 

2 

2 

2 

3 

.09 
.10 

1230 
1259 

1233 
1262 

1236 
1265 

1239 
1268 

1242 
1271 

1245 
1274 

1247 
1276 

1250 
1279 

1253 
1282 

1256 
1285 

0  1 
0  1 

1 

2 
2 

2 
2 

2 

3 

2 

3 

.11 

1288 

1291 

1294 

1297 

1300 

1303 

1306 

1309 

1312 

1315 

0  1 

2 

2 

2 

2 

3 

.12 

1318 

1321 

1324 

1327 

1330 

1334 

1337 

1340 

1343 

1346 

0  1 

2 

2 

2 

2 

3 

.13 

1349 

1352 

1355 

1358 

1361 

1365 

1368 

1371 

1374 

1377 

0  1 

2 

2 

2 

3 

3 

.14 

1380 

1384 

1387 

1390 

1393 

1396 

1400 

1403 

1406 

1409 

0  1 

2 

2 

2 

3 

3 

.15 

1413 

1416 

1419 

1422 

1426 

1429 

1432 

1435 

1439 

1442 

0  1 

2 

2 

2 

3 

3 

.16 

1445 

1449 

1452 

1455 

1459 

1462 

1466 

1469 

1472 

1476 

0  1 

2 

2 

2 

3 

3 

.17 

1479 

1483 

1486 

1489 

1493 

1496 

1500 

1503 

1507 1 1510 

0  1 

2 

2 

2 

3 

3 

.18 

1514 

1517 

1521 

1524 

1528 

1531 

1535 

1538 

1542' 1545 

0  1 

2 

2 

2 

3 

3 

.19 

1549 

1552 

1556 

1560 

1563 

1567 

1570 

1574 

1578  1581 

0  1 

2 

2 

3 

3 

3 

.20 

1585 

1589 

1592 

1596 

1600 

1603 

1607 

1611 

1614  1618 

0  1 

1 

2 

2 

3 

3 

3 

.21 

1622 

1626 

1629 

1633 

1637 

1641 

1644 

1648 

1652  1656 

0  1 

2 

2 

2 

3 

3 

3 

.22 

1660 

1663 

1667 

1671 

1675 

1679 

1683 

1687 

1690! 1694 

0  1 

2 

2 

2 

3 

3 

3 

.23 

1698 

1702 

1706 

1710 

1714 

1718 

1722 

1726 

1730  1734 

0  1 

2 

2 

2 

3 

3 

4 

.24 

1738 

1742 

1746 

1750 

1754 

1758 

1762 

1766 

1770  1774 

0  1 

2 

2 

2 

3 

3 

4 

.25 

1778 

1782 

1786 

1791 

1795 

1799 

1803 

1807 

1811  1816 

0  1 

2 

2 

2 

3 

3 

4 

.26 

1820 

1824 

1828 

1832 

1837 

1841 

1845 

1849 

1854  1858 

0  1 

2 

2 

3 

3 

3 

4 

.27 

1862 

1866 

1871 

1875 

1879 

1884 

1888 

1892 

1897  1901 

0  1 

2 

2 

3 

3 

3 

4 

.28 

1905 

1910 

1914 

1919 

1923 

1928 

1932 

1936 

1941 1 1945 

0  1 

2 

2 

3 

3 

4 

4 

.29 

1950 

1954 

1959 

1963 

1968 

1972 

1977 

1982 

1986; 1991 

0  1 

2 

2 

3 

3 

4 

4 

.30 

1995 

2000 

2004 

2009 

2014 

2018 

2023 

2028 

2032 

2037 

0  1 

2 

2 

3 

3 

4 

4 

.31 

2042 

2046 

2051 

2056 

2061 

2065 

2070 

2075 

2080 

2084 

0  1 

2 

2 

3 

3 

4 

4 

•32 

2089 

2094 

2099 

2104 

2109 

2113 

2118 

2123 

2128 

2133 

0  1 

2 

2 

3 

3 

4 

4 

.33 

2138 

2143 

2148 

2153 

2158 

2163 

2168 

2173 

2178 

2183 

0  1 

2 

2 

3 

3 

4 

4 

.34 

2188 

2193 

2198 

2203 

2208 

2213 

2218 

2223 

2228 

2234 

1  1 

2 

2 

3 

3 

4 

4 

5 

.35 

2239 

2244 

2249 

2254 

2259 

2265 

2270 

2275 

2280 

2286 

1  1 

2 

2 

3 

3 

4 

4 

5 

.36 

2291 

2296 

2301 

2307 

2312 

2317 

2323 

2328 

2333 

2339 

1  1 

2 

2 

3 

3 

4 

4 

5 

.37 

2344 

2350 

2355 

2360 

2366 

2371 

2377 

2382 

2388 

2393 

1  1 

2 

2 

3 

3 

4 

4 

5 

.38 

2399 

2404 

2410 

2415 

2421 

2427 

2432 

2438 

2443 

2449 

1  1 

2 

2 

3 

3 

4 

4 

5 

M 

2455 

2460 

2466 

2472 

2477 

2483 

2489 

2495 

2500 

2506 

1  1 

2 

2 

3 

3 

4 

5 

5 

.40 

2512 

2518 

2523 

2529 

2535 

2541 

2547 

2553 

2559 

2564 

1  1 

2 

2 

3 

4 

4 

5 

5 

.41 

2570 

2576 

2582 

2588 

2594 

2600 

2606 

2612 

2618 

2624 

1  I 

2 

2 

3 

4 

4 

5 

5 

.42 

2630 

2636 

2642 

2649 

2655 

2661 

2667 

2673 

2679 

2635 

1  1 

2 

2 

3 

4 

4 

5 

6 

.43 

2692  2698 

2704 

2710 

2716 

2723 

2729 

2735 

2742 

2748 

1  1 

2 

3 

3 

4 

4 

5 

6 

.44 

27541  2761 

2767 

2773 

2780 

2786 

2793 

2799 

2805 

2812 

1  1 

2 

3 

3 

4 

4 

5 

6 

.45 

2818  2825 

2831 

2838 

2844 

2851 

2858 

2864 

2871 

2877 

1  1 

2 

3 

3 

4 

5 

5 

6 

.46 

2884  2891 

2897 

2904 

2911 

2917 

2924 

2931 

2938 

2944 

1  1 

2 

3 

3 

4 

5 

5 

6 

47 

2951  2958 

2965 

2972 

2979 

2985 

2992 

2999 

3006 

3013 

1  1 

2 

3 

3 

4 

5 

5 

6 

.48 

3020  3027 

3034 

3041 

3048 

3055 

3062 

3069 

3076 

3083 

1  1 

2 

3 

4 

4 

5 

6 

6 

.49 1  30901  3097 

3105 

3112 

3119 

3126 

3133 

3141 

3148 

3155 

1  1 

2 

3 

4 

4 

5 

6 

_1 

ANTILOGARITHMS. 

1 

J.  a 

Proportional  Parts. 

II 

0 

1 

2 

3 

4 

5 

6 

7 

8 

9 

II 

1 

.50 

3162 

1 

1' 

I 

2 

4  5 

3 

6 

4  4 

7 
5 

8 
6 

9 
7 

3170 

3177 

3184 

3192 

3199 

3206 

3214 

3221 

3228 

.51 

3236 

3243 

3251 

3258 

3266 

3273 

3281 

3289 

3296 

3304 

2 

2 

3 

4  5 

5 

6 

7 

.52 

3311 

3319 

3327 

3334 

3342 

3350 

3357 

3365 

3373 

3381 

2 

2 

3 

4  5 

5 

6 

7 

.53 

3388 

3396 

3404 

3412 

3420 

3428 

3436 

3443 

3451 

3459 

2 

2 

3 

4  5 

6 

6 

7 

.54 

3467 

3475 

3483 

3491 

3499 

3508 

3516 

3524 

3532 

3540 

2 

2 

3 

4  5 

6 

6 

7 

.55 

3548 

3556 

3565 

3573 

3581 

3589 

3597 

3606 

3614 

3622 

2 

2 

3 

4  5 

6 

7 

7 

.56 

3631 

3639 

3648 

3656 

3664 

3673 

3681 

3690 

3698 

3707 

2 

3 

3 

4  5 

6 

7 

8 

.57 

3715 

3724 

3733 

3741 

3750 

3758 

3767 

3776 

3784 

3793 

2 

3 

3 

4  5 

6 

7 

8 

.58 

3802 

3811 

3819 

3828 

3837 

3846 

3855 

3864 

3873 

3882 

2 

3 

4 

4  5 

6 

7 

8 

.59 
.60 

3890 
3981 

3899 
3990 

3908 

3917 

3926 
4018 

3936 
4027 

3945 
4036 

3954 
4046 

3963 
4055 

3972 
4064 

2 
2 

3 
3 

4 
4 

5  5 
5  6 

6 
6 

7 
7 

8 
8 

3999 

4009 

.61 

4074 

4083 

4093 

4102 

4111 

4121 

4130 

4140 

4150 

4159 

2 

3 

4 

5  6 

7 

8 

9 

.62 

4169 

4178 

4188 

4198 

4207 

4217 

4227 

4236 

4246 

4256 

2 

3 

4 

5  6 

7 

8 

9 

.63 

4266 

4276 

4285 

4295 

4305 

4315 

4325 

4335 

4345 

4355 

2 

3 

4 

5  6 

7 

8 

9 

.64 

4365 

4375 

4385 

4395 

4406 

4416 

4426 

4436 

4446 

4457 

2 

3 

4 

5  6 

7 

8 

9 

.65 

4467 

4477 

4487 

4498 

4508 

4519 

4529 

4539 

4550 

4560 

2 

3 

4 

5  6 

7 

8 

9 

.66 

4571 

4581 

4592 

4603 

4613 

4624 

4634 

4645 

4656 

4667 

2 

3 

4 

5  6 

7 

9 

10 

.67 

4677 

4688 

4699 

4710 

4721 

4732 

4742 

4753 

4764 

4775 

2 

3 

4 

5  7 

8 

9 

10 

.68 

4786 

4797 

4808 

4819 

4831 

4842 

4853 

4864 

4875 

4887 

2 

3 

4 

6  7 

8 

9 

10 

.69 

4898 

4909 

4920 

4932 

4943 

4955 

4966 

4977 

4989 

5000 

2 

3 

5 

6;  7 

8 

9 

10 

.70 

5012 

5023 

5035 

5047 

5058 

5070 

5082 

5093 

5105 

5117 

2 

4 

5 

6^  7 

8 

9 

11 

.71 

5129 

5140 

5152 

5164 

5176 

5188 

5200 

5212 

5224 

5236 

2 

4 

5 

6  7 

8 

10 

U 

.72 

5248 

5260 

5272 

5284 

5297 

5309 

5321 

5333 

5346 

5358 

2 

4 

5 

6  7 

9 

10 

11 

.73 

5370 

5383 

5395 

5408 

5420 

5433 

5445 

5458 

5470 

5483 

3 

4 

5 

6  8 

9 

10 

11 

.74 

5495 

5508 

5521 

5534 

5546 

5559 

5572 

5585 

5598 

5610 

3 

4 

5 

6  8 

9 

10 

12 

.75 

5623 

5636 

5649 

5662 

5675 

5689 

5702 

5715 

5728 

5741 

3 

4 

5 

7j  8 

9 

10 

12 

.76 

5754 

5768 

5781 

5794 

5808 

5821 

5834 

5848 

586L 

5875 

3 

4 

5 

71  8 

9 

11 

12 

.77 

5888 

5902 

5916 

5929 

5943 

5957 

5970 

5984 

5998 

6012 

3 

4 

5 

7  8 

10 

11 

12 

.78 

6026 

6039 

6053 

6067 

6081 

6095 

6109 

6124 

6138 

6152 

3 

4 

6 

7  8 

10 

11 

13 

.79 

6166 

6180 

6194 

6209 

6223 

6237 

6252 

6266 

6281 

6295 

3 

4 

6 

7  9 

10 

11 

13 

.80 

6310 

6324 

6339 

6353 

6368 

6383 

6397 

6412 

6427 

6442 

1 

3 

4 

6 

7  9 

10 

12 

13 

.81 

6457 

6471 

6486 

6501 

6516 

6531 

6546 

6561 

6577 

6592 

2 

3 

5 

6 

8  9 

11 

12 

14  i 

•82 

6607 

662' 

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4 

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11 

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0569 

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4 

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23 

26 

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0828 

0864 

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0934 

0969 

1004 

1038 

107S 

1106 

3 

7 

10 

14 

17 

21 

24 

28 

31 

13 

1139 

1173 

1206 

1239 

1271 

1303 

1335 

1367 

13<l£ 

1430 

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16 

19 

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1461 

1492 

1523 

1553 

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1614 

1644 

1673 

170: 

1732 

3 

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18 

21 

24 

27 

15 

1761 

1790 

1818 

1847 

1875 

1903 

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1959 

1987 

2014 

3 

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8 

11 

14 

17 

20 

22 

25 

16 

2041 

2068 

2095 

2122 

2148 

2175 

2201 

2227 

2233 

2279 

3 

5 

8 

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13 

16 

18 

21 

24 

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2304 

2330 

2355 

2380 

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2430 

2455 

2480 

250-1 

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10 

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18 

2553 

2577 

2601 

2625 

264S 

2672 

2695 

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2742 

2765 

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2900 

2923 

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16 

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3010 

3032 

3054 

3075 

3096 

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3139 

3160 

3181 

3201 

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3263 

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3304 

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3345 

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2 

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10 

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3464 

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3747 

3766 

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2 

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9 

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3802 

3820 

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3874 

3892 

3909 

3927 

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2 

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14 

16 

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3979 

3997 

4014 

4031 

4048 

4065 

4082 

4099 

4116 

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9 

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4166 

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4200 

4216 

4232 

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11 

13 

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27 

4314 

4330 

4346 

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2 

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8 

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4713 

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9 

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5079 

5092 

5105 

5119 

5132 

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5198 

5211 

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5340 

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6405 

6415 

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2 

3 

4 

5 

6 

7 

8 

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6435 

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6454 

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6493 

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6561 

6571 

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2 

3 

4 

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6 

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8 

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6684 

6693 

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3 

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5 

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7 

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8 

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6730 

6739 

6749 

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6767 

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67S4 

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2 

3 

4 

5 

5 

6 

7 

8 

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6821 

6830 

6839 

6848 

6857 

6866 

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2 

3 

4 

4 

5 

6 

7 

8 

49 

6902 

6911 

6920 

6928 

6937 

6946 

6955 

6964 

6972 

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2 

3 

4 

4 

5 

6 

7 

8 

50 

6990 

6998 

7007 

7016 

7024 

7033 

7042 

7050 

7059 

7067 

2 

3 

3 

4 

5 

6 

7 

8 

51 

7076 

7084 

7093 

7101 

7110 

7118 

7126 

7135 

7143 

7152 

2 

3 

3 

4 

5 

6 

7 

8 

52 

7160 

7168 

7177 

7185 

7193 

7202 

7210 

7218 

7226 

7235 

2 

2 

3 

4 

5 

6 

7 

7 

53 

7243 

7251 

7259 

7267 

7275 

7284 

7292 

7300 

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7316 

2 

2 

3 

4 

5 

6 

6 

7 

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7332 

7340 

7348 

7356 

7364 

7372 

7380 

7388 

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2 

3 

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5 

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6 

7 

LOGARITHMS  OF  NUMBERS 

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0 

1 

2 

3 

4 

5 

6 

7 

8 

9 

Proportional  Parts. 

I 

2 

3 

4 

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6 

7 

8 

9 

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7419 

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7459 

7466 

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2 

2 

3 

4 

5 

5 

6 

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7490 

7497 

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7513 

7520 

7528 

7536 

7543 

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2 

2 

3 

4 

5 

5 

6 

7 

57 

7559 

7566 

7574 

7582 

7589 

7597 

7604 

7612 

7619 

7627 

2 

2 

3 

4 

5 

5 

6 

7 

58 

7634 

7642 

7649 

7657 

7664 

7672 

7679 

7686 

7694 

7701 

2 

3 

4 

4 

5 

6 

7 

59 

7709 

7716 

7723 

7731 

7738 

7745 

7752 

7760 

7767 

7774 

2 

3 

4 

4 

5 

6 

7 

60 

7782 

7789 

7796 

7803 

7810 

7818 

7825 

7832 

7839 

7846 

2 

3 

4 

4 

5 

6 

6 

61 

7853 

7860 

7868 

7875 

7882 

7889 

7896 

7903 

7910 

7917 

2 

3 

4 

4 

5 

6 

6 

62 

7924 

7931 

7938 

7945 

7952 

7959 

7966 

7973 

7980 

7987 

2 

3 

3 

4 

5 

6 

6 

63 

7993 

8000 

8007 

8014 

8021 

8028 

8035 

8041 

8048 

8055 

2 

3 

3 

4 

5 

5 

6 

64 

8062 

8069 

8075 

8032 

8089 

8096 

8102 

8109 

8116 

8122 

2 

3 

3 

4 

5 

5 

6 

65 

8129 

8136 

8142 

8149 

8156 

8162 

8169 

8176 

8182 

8189 

2 

3 

3 

4 

5 

5 

6 

66 

8195 

8202 

8209 

8215 

8222 

8228 

8235 

8241 

8248 

8254 

2 

3 

3 

4 

5 

5 

6 

67 

8261 

8267 

8274 

8280 

8287 

8293 

8299 

8306 

8312 

8319 

2 

3 

3 

4 

5 

5 

6 

68 

8325 

8331 

8338 

8344 

8351 

8357 

8363 

8370 

8376 

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2 

3 

3 

4 

4 

5 

6 

69 

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8401 

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8414 

8420 

8426 

8432 

8439 

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2 

3 

4 

4 

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6 

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8463 

8470 

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8488 

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2 

2 

3 

4 

4 

5 

6 

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2 

3 

4 

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2 

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4 

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73 

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2 

2 

3 

4 

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5 

74 

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8710 

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2 

2 

3 

4 

4 

5 

5 

75 

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8756 

8762 

8768 

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8779 

8785 

8791 

8797 

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2 

2 

3 

3 

4 

5 

5 

76 

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8814 

8820 

8825 

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8837 

8842 

8848 

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2 

2 

3 

3 

4 

5 

5 

77 

8865 

8871 

8876 

8882 

8887 

8893 

8899 

8904 

8910 

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2 

2 

3 

3 

4 

4 

5 

78 

8921 

8927 

8932 

8938 

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8960 

8965 

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2 

2 

3 

3 

4 

4 

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79 

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9004 

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2 

3 

3 

4 

4 

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80 

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3 

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81 

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9101 

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2 

3 

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4 

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82 

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9154 

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9165 

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9180 

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2 

2 

3 

3 

4 

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83 

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9212 

9217 

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9232 

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2 

2 

3 

3 

4 

5 

84 

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2 

2 

3 

3 

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85 

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9299 

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9315 

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2 

3 

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86 

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2 

2 

3 

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87 

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9400 

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9410 

9415 

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2 

2 

3 

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88 

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9460 

9465 

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2 

3 

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9571 

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2 

2 

3 

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91 

9590 

9595 

9600 

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9609 

9614 

9619 

9624 

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2 

2 

3 

3 

92 

9638 

9643 

9647 

9652 

9657 

9661 

9666 

9671 

9675 

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2 

2 

3 

3 

93 

9685 

9689 

9694 

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9703 

9708 

9713 

9717 

9722 

9727 

0 

2 

2 

3 

3 

94 

9731 

9736 

9741 

9745 

9750 

9754 

9759 

9763 

9768 

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2 

2 

3 

3 

95 

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9782 

9786 

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9795 

9800 

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9814 

9818 

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2 

2 

3 

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96 

9823 

9827 

9832 

9836 

9841 

9845 

9850 

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9859 

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0 

2 

2 

3 

3 

97 

9868 

9872 

9877 

9881 

9886 

9890 

9894 

9899 

9903 

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0 

2 

2 

3 

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98 

9912 

9917 

9921 

9926 

9930 

9934 

9939 

9943 

9948 

9952 

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2 

2 

3 

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99 

9956 

9961 

9965 

9969 

9974 

9978 

9983 

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9991 

9996 

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2 

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niH isdVHO IV  ON do AiisyaAiNn 


